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What is an Electron?
What is a proton?
Name the particles which determine the mass of an element.
What are α-particles?
What are the fundamental particles present in a neutral atom having atomic number greater than 1?
When α-particles are sent through a thin metal foil, most of them go straight through the foil. What inference do you draw from it?
What did Rutherford's experiment on scattering of particles show for the first time?
What is Plum-Pudding model of the atom?
J.J Thomson proposed that an atom possesses a spherical shape (radius approximately 10-10m) in which the positive charge is uniformly distributed. The electrons are embedded into it in such a manner as to give the most stable electrostatic arrangement. This model is known as Plum-pudding model of the atom.
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Why are electrons called planetary electrons?
What are nucleons?
When is the number of protons and electron same in an atom?
The number of electrons, protons and neutrons in a species are equal to 18, 16 and 16 respectively. Assign the proper symbol to the species.
The atomic number is equal to the number of protons = 16. The element is sulphur (S).
Mass No. = No. of protons + No. of neutrons
= 16 + 16 = 32.
Since the number of protons is not equal to the number of electrons, therefore the species is not neutral. It is anion (negatively charged) with charge equal to -2. Hence the symbol is 
.
Symbols 
can be written, whereas symbols 
are not acceptable. Answer Briefly.
. Thus, symbol 
is accrptable and
can be written but 
is not written. Since atomic number of an element is fixed, therefore it is not essential to indicate. However, mass number is not fixed or it depends upon the isotope taken. Hence it is essential to indicate mass number.
Atomic number is defined in terms of protons and not in terms of electrons. Why?
Calculate the number of neutrons in dipositive zinc ion (Zn2+, the mass number being 70).
No. of neutrons
= Mass number - Atomic number
Atomic mass of zinc is 70 g
Atomic number of Zinc is 30. Thus, number of neutrons will be,
No. of neutrons = 70 - 30 = 40
Calculate the number of electrons which together weigh one gram.
Mass of one electron =9.10939 x 10-31 kg
Therefore, Number of electrons that weigh 9.10939 x 10-31 kg =1
Number of electrons that will weigh1g= (1x10-3kg)
=0.1098 X10-3+31
=0.1098 X1028
=1.098 X1027
Find the number of electrons in K+ ion and Cl- ion (Z for K = 19, CI = 17).
Number of electrons in K+ = 19 - 1 = 18
Number of electrons in Cl- =17+1 = 18
What is the number of electrons, protons, neutrons and nucleons in carbon-14?
|
Planck |
Emission of light: through quanta. Propagation: through waves. |
|
Einstein |
Emission of light through photons. Propagation: through photons. |
Arrange the following type of radiations in increasing order of frequency:
(a) radiation from microwave oven
(b) amber light from traffic signal
(c) radiation from FM radio
(d) cosmic rays from outer space
(e) X-rays.
The increasing order of frequency is as follows:
FM radio< amber light<microwave<X-rays<cosmic rays
Give the expression for the angular momentum of a revolving electron in an atom.
Write the mathematical relation for determining the number of lines in the spectrum.
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where n is the outermost shell.
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What is the maximum number of lines when an excited electron of a hydrogen atom in n = 5 drops from exicted state to the ground state?
Hence, the maximum number of line is 10.
What is Bohr's radius?
Give an expression to calculate the radius of an orbit.
What is the region of light in which Lyman and Balmer series fall?
What are stationary states?
What is Zeeman effect?
It is the splitting up of spectral lines when atoms are subjected to a strong magnetic field.
What is Stark effect?
Calculate the radius of the fourth Bohr’s orbit of hydrogen atom.
Calcualtion of the radius of the fourth Bohr's orbit of hydrogen atom.
Radius of the orbits in hydrogen atom (r) given by,
n2 = x 0.529Å
∴ Radius of the fourth orbit
= (4)2 x 0.529A = 8.48Å
When white light that has passed through sodium vapours in viewed through a spectroscope, the observed spectrum has dark line at 5890Å. Explain this observation.
What is quantisation of energy?
An electron can emit or absorb energy as radiations only in fixed amounts or bundles called quanta. This is known as quantisation of energy.
Is an electron stationary in the stationary state?
What is the energy of an electron in the hydrogen atom in the ninth state?
For ninth excited state, n = 10. Thus,
Energy of an electron is -13.12kJ mol-1
What is the nature of electron, according to de-Broglie?
Give a mathematical relation between wavelength and momentum of a particle.
What is the most important application of de-Broglie concept?
Name the experiment used to verify the wave nature of electrons?
The wavelength of a proton, an electron and an α-particle moving with the same velocity are in the order: electron > proton > α-particle. How do you justify it?
Write the mathematical form of Heisenberg's uncertainty principle.
Heisenberg's mathmatical uncertainty principle.
where ∆x is uncertainty in position.
∆p is uncertainty in momentum.
∆v is uncertainty in velocity and h is
Planck’s constant
What is the meaning of orbital in terms of probability distribution of electrons?
What is the significance of de-Broglic concept and uncertainty principle in everyday life?
How is an electron identified in an atom?
Which quantum number specifies the shape of an orbital?
Write the possible values of m (magnetic quantum number) for an electron with given l (azimuthal quantum number)= 1.
When l = 1, then m = (2(l)+1) =3
the possible value of m is -1, 0, +1.
Which quantum number indicates the orientation of orbitals in space?
For 3d orbital. The principle quantum number is n=3.
(azimuthal quantum number) l = 2.
For the (magnetic quantum number ml) that is given by the rule such as (2(l) +1) = (2 x2 +1) =5. Thus, the possible value of ml = -2, -1, 0, +1 +2.
How many 3d electrons can have of spin quantum numbers = +1/2? What value of 1 these electrons will have?
How many electrons can be accommodated in a p-orbital?
Write the mathematical relation to calculate the number of electrons in the given subshell.
What designation is given to an orbital having:
(i) n = 3, l = 1 (ii) n = 4, l = 0?
For a particle atom, the maximum value of m obtained from experiment is +1. What are all the possible values of n, I and m?
What is meant by nodal plane in an orbital?
What is the shape of:
(i) an s-orbital (ii) a p-orbital?
(i) Spherically symmetrical and non-directional.
(ii) Dumb-bell and directional.
Which of the following orbitals is spherically symmetric?
(i) px (ii) s (iii) py
What is the shape of the orbital with:
(i) n = 2 and l = 0
(ii) n = 2 and l = 1?
(i) The orbital with n = 2 and l = 0 is orbital and has spherical shape.
(ii) The orbital with n = 2 and l = 1 is 2p orbital and has a dumb bell-shape.
What is (n+1) rule?
The (n+1) Rule is an empirical rule used to predict the multiplicity of electrons in shells of s, p,d, f orbitals.
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What is n(l)x notation?
n = principal quantum number
l =azimuthal quantum number
x = number of electrons.
Arrange the following orbitals in order of increasing energy:3s, 2p, 4d, 3p, 5s
How many electrons are there in the valence quantum level of copper(Atomic, number = 29)?
The atomic number of a certain element is 17. Determine the number of electrons in each shell of the atom.
Thus:
(i) K shell has 2 electrons.
(ii) L shell has 8 electrons.
(iii) M shell has 7 electrons.
What are n, l and m value for 2px and 3pz electrons?
(i) For 2px1 electron:
n-2, l= 1, m =-1
(ii) For 3pz electron:
n = 3, l= 1, m = + l
How many number of unpaired electrons are present in the ground state of Fe3+?(Atomic number of Fe is 26).
In which plane the probability of finding an electron residing in Px orbital is zero?
Write the electronic configuration of Cr3+. (At. No. of Cr = 24)
Properties of cathode rays:
1. Cathode rays travel in a straight line.
2. Cathode rays rotate the light paddle wheel when placed in their path. This shows that these rays consist of material particles having both mass and velocity.
3. Cathode rays produce green fluorescence upon striking glass or certain other materials.
4. When cathode rays are allowed to strike on a piece of metal foil, it becomes hot which shows that these rays produce heating effect.
5. These rays ionise the gas through which they pass.
6. These rays produce X-rays when they strike against hard metals such as tungsten.
7. When an electric or magnetic field is applied on the cathode rays, they are deflected from their straight path towards the positive plate of the electric field. This shows that cathode rays are negatively charged particles. These particles were named electrons.
How can you say that electron is a universal constituent of all atoms?
Discharge tube experiments have revealed that electrons constituting the cathode rays are identical irrespective of material of cathode or gas used in the discharge tube. All these electrons are found to have same value of e/m . Thus cathode rays consist of fundamental common particles known as electrons.
Moreover, electrons can be also emitted:
(i) by heating certain metal filaments to high temperature,
(ii) by the radioactive substances in the form of β-rays and
(iii) by exposing certain metal surfaces especially the alkali metals to certain suitable high frequency radiations such as ultraviolet or X-rays.
From the above discussion, it is clear that electrons can be emitted from all kinds of matter and value of e/m is found to be the same regardless of the source or method by which electrons are obtained. Therefore, it can be concluded that electrons are universal (or common) constituents of all matter.
What are anode rays? How do they originate? Briefly explain their important properties.
The streams of positively charged particles are called anode rays or positive rays or canal rays.
In 1886, Goldstein performed a discharge tube experiment, using perforated cathode. He observed that in addition to cathode rays, a new kind of rays were also found. These rays passed through the hole of the perforated cathode but travelled in a direction opposite to that of cathode rays. These rays were found to consist of positively charged particles and were called anode rays or positive rays or canal rays.
These rays are believed to be produced as a result of the knock out of the electrons from the gaseous atoms by the bombardment of high speed electrons of the cathode rays on them. Thus(anode rays are not emitted from the anode but are produced in the space between the anode and the cathode. In other words anode rays (or positive rays) are atomic or molecular residues from which some electrons have been removed. The removed electrons constitute the cathode rays and positive residues form the positive or canal rays.
Properties of anode rays:
(i) Anode rays travel in a straight line.
(ii) These rays are deflected by electric and magnetic fields in a way that shows that these rays are positively charged particles. For example, these rays are attracted towards the negative plate in the electric field which means rays are positively charged particles.
(iii) These rays consist of material particles and rotate the paddle wheel placed in their path.
(iv) Anode rays produce heating effect when struck against a metal foil.
(v) Hie ratio of charge to mass (e/m) for positive rays is considerably smaller than for electrons and is not constant but depends on the nature of the gas taken in the discharge tube. In other words, the value of e/m for a positive ion depends upon the charge of the ion and its mass.
What is the difference in the origin of cathode rays and anode rays?
Cathode rays consist of negatively charged material particles called electrons. These rays deflect towards the positive plate of an electric field. The charge to mass (e/m) for the particles in the nature of the gas taken in the discharge tube.
Anode rays consists of positively charges material particles. These rays deflect towards the negative plate of an electric field.
Describe briefly Thomson model of an atom.
How is position of electron and proton in an atom established? Describe an experiment in this regard.
Or
Describe Rutherford's scattering experiment. How did it lead to Rutherford’s nuclear model of atom?
Or
Describe an experiment which provides evidence that most of the charge and mass of an atom is concentrated in its nucleus.
On the basis of the above observations about the scattering of alpha particles, he proposed a model of the atom called Rutherford model of the atom.
Rutherford’s atomic model. According to this model:
1. Almost the entire mass and all of the positive charge of an atom is concentrated in a very small region at the centre of the atom known as a nucleus.
2. The size (diameter) of the nucleus (nearly 10-13 cm) is very small as compared to the size of the atom (nearly 10-8cm).
How will you explain the observations of Rutherford's α-ray scattering experiment?
(i) Most of the space in an atom between the nucleus and revolving electrons is empty. So most of α-particles will pass through an atom without any deflection. Electrons may cause only minor deflections to α-particles.
(ii) Alpha particles (He2+) have two units of positive charge while the nucleus of gold contains 79 protons. The nucleus will repel the α-particles. Since the size of the nucleus is very small, so only a few α-particles will pass by its side and deflected by a small angle.
(iii) A very few α-particles (1 in 20000) may hit or try to hit the small nucleus and these will be deflected backwards through angles greater than 90°.
Discuss the drawbacks or limitations of Rutherford model of atom.
According to Rutherford’s model, electrons are revolving around the nucleus in circular orbits. The centrifugal force (due to the circular motion of electrons) acting outwards balances the electrostatic force of attraction (between the positively charged nucleus and negatively charged electrons) acting inwards. This prevents the electrons to fall into the nucleus.
It was shown by Clark Maxwell that a charged body moving under the influence of attractive force loses energy constantly. Thus, unlike a planet, electron is a charged particle and it should continuously emit radiation and lose energy. As a result of this, a moving electron will come closer and closer to the nucleus and after passing through a spiral path, it should ultimately fall into the nucleus. It has been calculated that it should take only 10-8 sec for the electron to fall into the nucleus. But it is known that electrons keep on moving outside the nucleus. Hence there must be something wrong with Rutherford's atomic model itself.
This model also fails to explain hydrogen spectrum.
Fig. The orbit becoming smaller and smaller as electron comes closer to the nucleus through spiral path
How was neutron discovered? Describe briefly.
It was Rutherford who showed that almost the entire mass of an atom is concentrated in the nucleus. However, he could not explain the total known mass of the atom in terms of the number of protons in the nucleus. For example, helium nucleus has a charge of 2+ which is furnished by 2 protons, also contributes approximately 2 units of mass. As the mass of helium atom is 4 a.m.u., therefore 2 units of mass are still to be accounted for.
Rutherford postulated the existence of a neutral particle to account for the missing mass. Chadwick in 1932, actually discovered this neutral particle by bombardment of beryllium or boron by alpha particles. It has almost exactly the same mass as of a proton but carried no electric charge. These neutral particles were called neutrons.
Thus, a neutron is a neutral particle having a mass (1.675 x 10-27 kg) which is almost exactly the same as that of a proton. On the a.m.u. scale, the mass of the neutron is 1.00867.
What are atomic number and mass number? How are they related to each other?
Atomic number (Z): In 1913, Mosley devised an experiment to find the positive charge on the nucleus of an atom. Atomic number of an element is equal to the number of unit positive charges or protons present in the nucleus of an atom of that element.
An atom is electrically neutral, therefore, in an atom,
Number of protons = No. of electrons
Atomic number = No. of protons in the nucleus of the atom
= No. of electrons present outside the nucleus of the same atom.
Atomic number is a fundamental property of the atom and is generally denoted by Z.
Mass number (A):The mass of an atom is mainly concentrated in the nucleus. In the nucleus, there are protons and neutrons, these are collectively called nucleons. Mass number of an element is the sum of the number of protons and neutrons present in its atom. It is generally denoted by A.
Mass No. (A) = No. of protons + No. of neutrons = No. of nucleons.
Relation between mass number and atomic number:
We know,
Mass number (A) = No. of protons + No. of neutrons
= At. No. (Z) + No. of neutrons.
No. of neutrons = Mass No. (A) - Atomic No. (Z)
Calculate the mass and charge of one mole of electrons.
One mole of electron
Mass of 1 electron = 
Mass of 
Charge on one electron
Charge on one mole electrons
Here number of protons = 13
Number of electrons = 13
Atomic number = Number of protons = 13
Number of neutrons = Mass No. (A) - At. No. (Z) = 27- 13= 14
Give the number of electrons in the species 
No. of electrons in 
Number of electrons in 
Number of electrons in 
Write the complete symbol for the atom with the given atomic number (Z) and atomic number (A).
(i) Z = 17, A = 35
(ii) Z = 92, A = 233
(iii) Z = 4, A = 9.
(i) The element with atomic number 17 is Cl. Its symbol is 
(ii) The element with atomic number 92 is U. Its symbol is 
(iii) The element with atomic number 4 is Be. Its symbol is 
No. of protons + No. of neutrons = Mass number = 81
Let number of protons in the atom be x
Atomic symbol is 
Number of protons = x -1
Symbol of the ion is 
-.
Symbol of the ion is 
What is the atomic number and mass number of A3+ ion with 21 electrons and 25 neutrons?
Number of protons = 21 + 3 = 24
Atomic number = 24
Mass number = No. of protons + No. of neutrons
= 24 + 25 = 49.
Define isotopes, isobars and isotones.
Isotopes: Isotopes are atoms of the same element having same atomic number but different mass numbers. e.g. 
Isobars: Isobars are atoms of different elements having different atomic numbers but the same mass number. e.g. 
are isobars of each other.
Isotones: Isotones are atoms of different elements having same number of neutrons. For example 
Name different isotopes of hydrogen. Give their atomic numbers and mass numbers.
Isotopes of an element have similar chemical properties and isobars have different chemical properties. Explain.
Isotopes of an element have same number of electrons and protons but have different number of neutrons. Since chemical properties of elements depend upon atomic number, isotopes of an element have similar chemical properties.
On the other hand, isobars have different number of electrons, protons as well as neutrons. Thus, isobars have different chemical properties.
If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by side in a straight line across length of scale of length 20 cm long.
Diameter of carbon atom = 0.15 nm
Length along which carbon atoms are to be placed
= 20 cm
Hence, number of carbon atoms which can be placed along the length
We have given,
Length of the given arrangement =2.4 cm
Number of carbon atoms present =2 x 108
Therefore, diameter of carbon atom
= 2.4x10-2 m/2 x108 =1.2 x10-10m
Thus, radius of carbon atom = Diameter/2
=1.2 x10-10/2 =6.0 x10-11m
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A certain particle carries 2.5 X 10-16C of static electric charge. Calculate the number of electrons present in it.
We have given,
Number of electron carrying charge of 2.5 x 10-16C
Charge on one electron =1.6022 x10-19C
1.6022 x10-19C charge is carried by 1 electron,
= 1 x (2.5 x10-16C) /1.6022 x10-19C
=1.560 x103C
=1560C
In Milikan's experiment, static electric charge on the oil drops has been obtained by shining X-rays. If the static electric charge on the oil drop is - 1.282 X 10-18C, calculate the number of electrons present on it.
Charge on the oil drop =1.2820x10-18C
Charge on the electron =1.6022 x10-19C
Therefore,
Calculate the total number of electrons present in one mole of methane.
1 molecule of 
contains electrons = 6 + 4 = 10
(a) 1 g atom of carbon = 14g
= 6.022 x 1023 atoms
Each carbon atom 
(b) Mass of 1 neutron = 1.675 x 10-27 kg
Mass of 2.4088 x 1021 neutrons
= (1.675 x 10-27kg) x (2.4088 x 1021)
= 4.0347 x 10-6kg
Find: (a) the total number and (b) the total mass of protons in 34 mg of NH3 at STP.
Will the answer change if the temperature and pressure are changed ?
(a) 1 molecule of NH3 contains
= 7 + 3 = 10 protons
Now 17g NH3 contains 6.022 x 1024 protons
(b) Mass of one proton = 1.6726 x 10-27 kg
There is no effect of temperature and pressure.
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Light is an important form of energy, According to Newton's corpuscular theory,the light was considered as a stream of particles called corpuscles of light. This theory could explain the phenomenon of reflection and refraction, but failed to explain the phenomenon of diffraction and interference. Therefore, the corpuscular theory was discarded and replaced by the wave theory of light. According to this theory, light is transmitted in the form of electromagnetic waves. These waves are associated with the oscillating electric field and magnetic fields.
The energy emitted from any source in the form of radiation is called radiant energy. The most common forms of radiant energy are cosmic rays, gamma rays, X-rays, ultraviolet rays, visible rays, infrared rays, radio waves and heat. Clark Maxwell (1856) proposed that radiant energy has wave character. These waves are associated with both electric and magnetic fields, thus, these radiations are called electromagnetic radiations. All types of electromagnetic radiations travel through space with the same velocity i.e. the velocity of light (3 x 108 ms-1).
What are the characteristics of waves?
The following parameters are used to characterise these waves:
(i) Wavelength: It is defined as the distance between two adjacent crests or troughs of the wave. It is denoted by lambda (λ). It is generally expressed in metres; Angstroms (Å) or nanometres (nm).
1 Å = 10-10m
1 nm = 10-9 m
(ii) Frequency: It is defined as number of waves which pass through a given point in one second. It is denoted by v (nu) and is measured in cycles per second (cps) or simply reciprocal seconds (sec-1). It is also expressed in terms of Hertz (Hz).
It is inversely proportional to the wavelength.
(iii) Velocity: The distance travelled by the wave in one second is called the velocity of the wave. It is denoted by c and is equal to the product of wavelength and frequency.
c =vλ
The units of velocity are ms-1 or cm s-1.
Light and all other types of electromagnetic radiations travel through space with the same velocity i.e. 3 x 10-8 ms-1.
(iv) Amplitude: The height of the crest or the depth of the trough or distance of maximum displacement is called amplitude. It is denoted by a. It determines the density or brightness of a beam of light.
(v) Wave number: It is defined as the number of wavelengths per unit of length and is equal to the inverse of wavelength expressed in centimetres. It is denoted by ⊽ cm-1 or in-1.
...(1)
But 
or 
...(2)
From (1) and (2),
or 
Calculate the frequency and wavelength of light of wavelength:
(i) 5000 A (ii) 4.4 mμ (iii) 580 mm
(i) Wavelength of light, 
Velocity of light, 
Wave number,
(ii) Wavelength of light,
Velocity of light c = 3 x 108 ms-1
(iii) Wavelength of light, 
= 580 nm
= 580 x 10-9m
Velocity of light, c = 3 x 108 ms-1
Here c = 3.00 x 108 ms-1;
This is the characteristic radio wave-length.
A certain radiostation broadcasts on a frequency of 920 k Hz. What is the wavelength of electromagnetic radiation broadcast by the radio station?
The frequency of radiostation emitted.
v = 920 kHz = 920 x 103 Hz = 920 x 103 s-1
We know, 
or 
Wavelength of radiowave broadcast by the station = 326 m
The frequency of D-line in the spectrum of sodium is 5.09 x 1014s-1. Calculate the wavelength of light.
Frequency of radiation
= 5.09 x 1014 s-1
Velocity of light = 3 x 108 ms-1
We know, 
or 
The wavelength range of the visible spectrum extends from violet (400 nm) to red (750 nm). Express these wavelengths in frequencies (Hz). [1 nm = 10-9 m]
Frequency of violet light,
Frequency of red light,
What is continuous spectrum?
What is an emission spectra? What are its types?
When the radiation emitted from some source e.g. from the sun or by passing electric discharge through a gas at low pressure or by heating some substance to high temperature etc. is passed directly through the prism and recorded over a photographic plate or screen, the spectrum obtained is known as an emission spectrum.
Types of an emission spectrum. Emission spectra can be further divided into three types:
(i) Continuous spectrum. When white light from the sun or an electric bulb is analysed by passing through a prism, the spectrum obtained consists of a continuous band of seven colours from red to violet (like a rainbow). Such a spectrum is known as continuous spectrum. In such a spectrum, there is a continuous sequence of impressions in the form of colours.
(ii) Line spectrum. When the radiations emitted by the excited substances e.g. by placing some volatile salt (say NaCl) in the bunsen flame or by
passing an electric discharge through a gas at low pressure are analysed by a spectrograph, discontinuous spectra consisting of a series of sharp lines and separated by dark bands is obtained. Such a spectrum consisting of parallel lines separated by dark space is called line spectrum. Line spectra is a finger -print of atoms because :
(a) each element has its characteristic spectrum, differing from those of all other elements and
(b) each line in the spectrum corresponds to a particular wavelength.
(iii) band spectrum. A molecule (unlike the atom) emits the radiations over a close range of wavelengths so that the spectrum consists of the group of lines spaced very close together. These lines are so much close that they look like a band. Such a spectrum is known as band spectrum.
What do you mean by absorption spectra?
Hydrogen spectrum is an important example of line spectrum. The spectrum of hydrogen can be obtained by passing an electric discharge through the hydrogen gas taken in the discharge tube under low pressure. The radiation thus obtained is analysed with the help of a spectroscope The spectrum thus obtained contains a large number of lines present in the ultraviolet, visible and infra-red regions. These lines have been grouped into five series, each named after the name of the discoverer.
These are:
(i) Lyman series..........Ultraviolet region
(ii) Balmer series..........Visible region
(iii) Paschen series
(iv) Brackett series.........Infrared series
Though hydrogen atom contains only one electron but so many lines are obtained in the spectrum.
It is because of different possibilities of its excitation in its different energy levels. The lines in the visible region can be directly seen on the photographic film while the position of lines in the ultraviolet and infrared regions are located employing different techniques.
Describe briefly Planck's Quantum Theory.
This theory can be simply stated as:
(i) Radiation is associated with energy.
(ii) Radiant energy is not emitted or absorbed continuously but discontinuously in the form of small packets of energy.
(iii) Each packet of energy is associated with a definite amount of energy and is known as quantum. In the case of light, the ‘quantum’ is known as ‘photon’.
(iv) The magnitude of quantum is directly proportional to the frequency of radiation.
E ∝ v
i.e. E = hv
where v = Frequency of radiation
h = Planck’s constant
= 6.625 x 10-34Js
(or 6.625 X 10-34 kg m2 s-1)
= 3.99 x 10-13 k Js mol-1
(v) A body can radiate or absorb energy in whole number multiples of a quantum i.e. nhv, where n = 1,2,3,4......etc.
Explain briefly the photoelectric effect.
The phenomenon of ejection of electrons from the surface of a metal when the light of suitable frequency strikes it is called photoelectric effect and the emitted electrons are called photoelectrons.
For each metal, there is a characteristic minimum frequency called threshold frequency below which the photoelectric effect does not occur. For the photoelectric effect to occur, the striking photon should have frequency more than that of the threshold frequency. If a photon of frequency v strikes a metal atom whose threshold energy is v0, then photoelectrons will be emitted only if v > v0. Since the striking photon has energy equal to hv and minimum energy required to eject electron is hv0 (called wave function W0), then the excess of energy i.e. hv - hv0 or h(v - v0) will be imparted to the ejected electron as kinetic energy.
Hence,
K.E. of ejected electron 
where me is the mass of electron and v be its velocity.
A photon of wavelength 4 x 10-7 m strikes on metal surface, the work function of the metal being 2.13 eV. Calculate: (i) the energy of the photon (eV), (ii) the kinetic energy of the emission, and (iii) the velocity of the photoelectron (1 eV = 1.6020 10-19J).
(i) Eneryg of one photon 
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= hv - hv0 = 3.10 - 2.13 = 0.97 eV
(iii) 
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Wavelength of radiation (λ) = 6800 Å
= 6800 X 10-10m
Now, threshold frequency 
Also work function (W0) = hv0
= (6.626 x 10-34 Js) x (4.41 x 1014 s-1)
= 2.92 x 10-19 J
The work function for caesium atom is 1.9 eV. Calculate: (a) the threshold wavelength and (b) the threshold frequency of radiation. If the caesium element is irradiated with a wavelength 500 nm, calculate the kinetic energy and velocity of the ejected photoelectron.
Work function 
...(1)
We know, 
Substituting the values in eq. (2), we have
We know
Energy of the incident radiation = Work function + KE of photoelectron
Work function = Energy of the incident
radiation - KE of photoelectron ..(1).
Now energy of the incident radiation (E) = hv
..(2)
Substituting the values in eq. (2), we have
Energy of incident radiation
The potential applied gives the kinetic energy to the electron.
Hence, the kinetic energy of the electron = 4.4 eV.
Substituting the values in eq. (1), we have
Work function = 4.83 eV - 0.35 eV
= 4.48 eV
Application of Quantum theory:
(i) In 1905, Einstein applied this theory to explain the photoelectric effect.
(ii) In 1913, N. Bohr used this theory to explain the structure of atoms and explain hydrogen spectra.
(iii) In 1922, this theory was used to explain Compton effect.
Find energy of each of the photons which correspond to light of frequency 3 x 1015 Hz.
Here v = 3 x 1015 Hz.
But E = hv
Substituting the values, we have,
Calcualte the energy of one mole of photons of radiation whose frequency is 5 x 1014Hz.
Energy (E) of one photon = hv
Here, h = 6.626 x 10-34 Js
v = 5 x 1014 s-1
Energy of one photon of light (E) = hv
Energy of n photons of light (En) = nhv
or 
or 
...(1)
Here, 
Substituting the value in eq. (1), we have,
A 100 watt bulb emits monochromatic light of wavelength 400 nm. Calculate the number of photons emitted per second by the bulb.
Power of the bulb = 100 watt
= 100 Js-1
Energy of one photon E = hv = 
Power of the bulb = 25 watt = 25 Js-1
Energy of one photon 
or
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Nitrogen laser produces a radiation at a wavelength of 337.1 nm. If the number of photons emitted is 5.6 x 1024, calculate the power of this laser.
Energy of one photon of light (E) = hv
Energy of n photon of light (En) = nhv
...(1)
Here, n = 5.6 x 1024; h = 6.626 x 10-34 Js;
Substituting the values in eq. (1), we have
Neon gas is generally used in the sign boards. If it emits strongly at 616 nm, calculate: (a) the frequency of emission, (b) distance travelled by this radiation in 30 s (c) energy of quantum and (d) number of quanta present if it produces 2 J of energy.
Here, 
(a) Frequency of emission,
(b) Velocity of radiation = 
(c) 
(d) Number of quanta in 2J of energy
Here, 
According to Planck's quantum theory,
....(1)
Substituting the value in (1),
The energy (E) associated with a 300 nm photon is given by
E = hv
The minium energy needed to remove a mole of electrons from sodium
The minimum energy for one electron
Energy (E) of one photon = hv = 
...(1)
Here 
Substituting the values in eq. (1), we have
Lifetimes of the molecules in the excited states are often measured by using pulsed radiation source of duration nearly in the nano second range. If the radiation source has the duration of 2 ns and the number of photons emitted during the pulse source is 2.5 x 1015, calculate the energy of the source.
Frequency = 
Energy = nhv
Electromagnetic radiation of wavelength 285 nm is just sufficient to ionise the potassium atom. What is the ionization energy of potassium (in kJ mol-1) ?
...(1)
Energy of one photon (E) = hv
...(1)
This energy of a photon is just sufficient to ionise a sodium atom.
Thus ionisation energy of sodium
Give essential features of Bohr's model of atom.
2. These circular paths are called orbits, shells energy levels or stationary states in which electron can revolve around the nucleus without emitting radiations. So far as an electron revolves in a certain orbit, its energy remains constant.
3. These different energy levels are designated by numbers 1, 2, 3, 4 etc. or letters K, L, M, N etc. starting from the nucleus. The greater the distance of the energy level from the nucleus more is the energy associated with it.
4. The electrons in an atom can revolve only in those orbits in which the angular momentum (mvr) of the electron is a whole number (n) multiple of a constant 
.
where m = mass of electron,
v = velocity of electron
r = radius of orbit.
h = Planck's constant,
n = whole number
This postulate indicates that the angular momentum of an electron moving in a circular orbit is quantisied. The angular momentum can be
5. When an electron jumps from one stationary state to another, the difference of energy (∆E) between two states (E1 and E2) is emitted or absorbed, as radiation of frequency (v) given by the equation
∆E = E2- E1 = hv
If an electron jumps from higher energy state to a lower energy state, energy is emitted. Energy is absorbed by an electron when it jumps from a lower energy state to a higher energy state.
It is obvious that the electron cannot radiate energy if no energy level is available. That is why atoms do not collapse.
From Bohr model, one can calculate the energy En of an electron in an orbit n. This is given by the expression,
Further, one can also calculate the radius of each circular orbit from the expression
rn = 0.529Å x n2 where n= 1,2,3......
The radius of the first orbit r1, called Bohr’s radius (n = 1) is 0.529A (or 52.9 pm).
Bohr model is also applicable to ions such as He+, Li2+ etc. For such cases,
and 
where Z is the atomic number and has values of 2 and 3 for He+ and Li2+ respectively.
The energy associated with the first orbit in the hydrogen atom is - 2.18 x 10-18J atom-1. What is the energy associated with the fifth orbit?
(ii) Calculate the radius of Bohr’s fifth orbit for the hydrogen atom.
ii) Radius of Bohr’s nth orbit for hydrogen atom is given by,
rn =(0.0529)n2
For
n=5
rs =0.0529nm x 5 x5
rs =1.3225nm
Calculate the energy associated with first orbit of He+. What is the radius of this orbit?
We know,
Here n = 1: Z = 2 for He+
Also we know,
How much energy is required to ionise a H atom if the electron occupies n = 5 orbit? Compare your answer with the ionization enthalpy of H atom (energy required to remove the electron from n = 1 orbit).
Let the threshold wavelength = λ0nm
= λ0 x 10-9m
Then 
or 
For the first experiment:
For seconds experiment:
For the third experiment:
...(3)
Dividing equation (2) by equation (1), we have
Substituting this value in equation (3), we have
How does Bohr's model explain the stability of an atom?
How does Bohr's model of atom explain the atomic spectra of hydrogen and hydrogenlike particles?
There is only one electron in hydrogen atom but the hydrogen spectrum consists of a large number of lines in various regions of radiations namely ultraviolet, visible and infra-red. Bohr’s atomic theory provides a satisfactory explanation for the emission of atomic spectra of atoms containing one electron i.e. H, He+ , Li2+ etc.
When an electric discharge is passed through a tube containing hydrogen gas at low pressure, hydrogen molecules dissociate to form hydrogen atoms. These hydrogen atoms absorb energy and the electrons in them are promoted to higher energy levels from their ground state (n = 1). Since in a sample of hydrogen, there are large number of atoms, the electrons in different atoms absorb different amounts (quanta or photons) of energies and are accordingly promoted to different energy states (2, 3, 4, 5....)
All these excited states are metastable states. The electrons cannot remain in these forever. They soon radiate energy and return back to the ground state (n = 1) and others n = 2, n=3, n=4 etc. The electron may return to the lower states in one or more jumps. These transitions emit radiations of different frequencies or wave numbers and produce different lines in the hydrogen spectrum.
The difference in energy between two energy levels is related to the frequency of the radiation emitted as:
or 
where E2 and E1 represent the energies of the higher and lower energy levels respectively. ∆E is the difference in their energies, v is the frequency and h are Planck’s constant.
Thus, every line in the hydrogen spectrum corresponds to a particular drop from some higher to some lower energy level as shown.
The lines in the Lyman series are obtained when electrons drop from higher energy levels (i.e n = 2, 3, 4 etc.) to the first energy level (i.e. n =1). These lines fall in the ultraviolet region.
The lines in the Balmer series are obtained when electrons drop from higher energy levels (i.e. n = 3. 4, 5, 6 etc.) to the second energy level (i.e. n =2). These lines fall in the visible region.
Similarly, lines in Paschen, Brackett and Pfund series are obtained when electrons drop from higher energy levels to the third (n =3), fourth (n=4) and 5th energy level (n = 5) respectively. These lines fall in the infra-red region. In short,
Lyman series From n = 2, 3, 4, 5,6 .... to n = 1
Balmer series From n = 3,4, 5, 6 .... to n = 2
Paschen series From n = 4, 5,6..... to n = 3
Brackett series From n = 5,6, 7..... to n = 4
Pfund series From n = 6,7....... to n = 5
Derive the expressions:
What are frequency and wavelength of a photon emitted during a transition from n=5 state to n = 2 state in hydrogen atom?
What is the energy in joules, required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit and what is the wavelength of the light emitted when the electron returns to the ground state? The ground state electron energy is - 2.18 X 10-11 ergs
As ground state, electronic energy is 
When electron returns to ground state (i.e. to n = 1), energy emitted 
We know,
Substituting the values in the above expression
What are the limitations (drawback or shortcomings) of Bohr's model of atom?
The main limitations of Bohr’s model are:
1. Bohr’s theory could successfully explain hydrogen spectrum and spectra of single electron species, but it failed to explain the spectra of multielectron atoms (atoms containing more than one electron).
2. Bohr’s theory failed to explain the splitting of spectral lines into a group of finer lines under the influence of a magnetic field (Zeeman effect) and of an electrical field (Stark effect).
3. Bohr’s theory failed to explain the motion of an electron in an atom in three dimensions.
4. Bohr’s theory could not explain the shapes of molecules formed by the combination of atoms.
5. de-Broglie (in 1923) suggested that electron has a dual character. It behaves like a particle as well as like a wave. But Bohr had treated electron only as a particle.
6. Bohr’s theory assumes that an electron revolves around the nucleus at a certain fixed distance from the nucleus with a definite velocity. This is not in accordance with Heisenberg’s uncertainty principle.
What is Rydberg formula? What is the value of Rydberg constant ? What is its significance?
Calculate the wavelength of the first and limiting spectral line in the Lymann series of hydrogen spectrum.
Calculate the wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen.
,
What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of He+ spectrum?
Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represented as
v = 3.29 x 1015 (Hz) [1/32- l/n2.]
Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum.
Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum.
We know,
Radius of nth orbit of H-like partiles
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#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(112): com_wiris_util_sys_Store->write('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
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#7 {main}</pre>](/application/zrc/images/qvar/CHEN11085277.png)
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/sys/io/File.class.php(12): file_put_contents('/home/config_ad...', 'mml=<math xmlns...')
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(48): sys_io_File::saveContent('/home/config_ad...', 'mml=<math xmlns...')
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(112): com_wiris_util_sys_Store->write('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
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#7 {main}</pre>](/application/zrc/images/qvar/CHEN11085277-1.png)
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#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(48): sys_io_File::saveContent('/home/config_ad...', 'mml=<math xmlns...')
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(112): com_wiris_util_sys_Store->write('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
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#7 {main}</pre>](/application/zrc/images/qvar/CHEN11085277-2.png)
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#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(112): com_wiris_util_sys_Store->write('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
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Dividing equation (1) by equation (2)
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#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(48): sys_io_File::saveContent('/home/config_ad...', 'mml=<math xmlns...')
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#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
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#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(48): sys_io_File::saveContent('/home/config_ad...', 'mml=<math xmlns...')
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(112): com_wiris_util_sys_Store->write('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
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#7 {main}</pre>](/application/zrc/images/qvar/CHEN11085277-5.png)
Why the energy of electron in an atom is negative?
What do you understand by Dual nature of matter?
Or
What is de-Broglie relationship?
In 1905, Einstein suggested that lights has wave nature and particle nature(dual nature). In 1923, de-Broglie proposed that like light, matter (all materials particles)a lso has wave as well as particle nature(dual matter). He derived a relationship for calculation of the wavelength (A) of the wave associated with a particle of mass m, moving with a velocity v.
The above relation is known as de-Broglie’s relation and states that the wavelength of the matter wave is inversely proportional to its momentum. In 1927, Davisson and Germer verified experimentally the wave nature of electron.
How will you correlate Bohr’s concept of hydrogen atom with de-Broglie concept?
Or
Show that the circumference of Bohr orbit for the hydrogen atom is an integral multiple of de-Broglie wavelength associated with electron revolving around the orbit.
...(2)
...(3)What will be the wavelength of a ball of mass 0.1 kg moving with a velocity of 10 ms-1?
If the velocity of the electron in Bohr's first orbit is 2.19 x 106 ms-1, calculate the de Broglie wavelength associated with it.
....1
Dual behaviour of matter proposed by de Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules and other type of material. If the velocity of the leectron in this microscope is 1.6 X 106 ms-1, calculate de Broglie wavelength associated with this electron.
...(1)
According to de-Broglie equation,
Substituting the value in eq. (1), we have,
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/sys/io/File.class.php(12): file_put_contents('/home/config_ad...', 'mml=<math xmlns...')
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(48): sys_io_File::saveContent('/home/config_ad...', 'mml=<math xmlns...')
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(112): com_wiris_util_sys_Store->write('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#6 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#7 {main}</pre>](/application/zrc/images/qvar/CHEN11085342.png)
is constant, this means that the two are inversely proportional to each other. Thus if position is known with more accuracy, there will be more uncertainty in momentum and vice versa.
Why is Heisenberg’s uncertainty principle not useful in daily life?
A golf ball has a mass of 40g and a speed of 45m/s. If the speed can be measured within accuracy of 2%, calculate the uncertainty in the position.
If the position of the electron is measured within an accuracy of ![<pre>uncaught exception: <b>file_put_contents(/home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/web/../../../../../../formulas/23/fe/1ee5931355f67d0ed4b6276908fd.ini): failed to open stream: Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/sys/io/File.class.php at line #12file_put_contents(/home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/web/../../../../../../formulas/23/fe/1ee5931355f67d0ed4b6276908fd.ini): failed to open stream: Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/sys/io/File.class.php line 12<br />#0 [internal function]: _hx_error_handler(2, 'file_put_conten...', '/home/config_ad...', 12, Array)
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calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is![<pre>uncaught exception: <b>file_put_contents(/home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/web/../../../../../../formulas/43/3b/2b0f3da5ba43fde12258af6b35c2.ini): failed to open stream: Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/sys/io/File.class.php at line #12file_put_contents(/home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/web/../../../../../../formulas/43/3b/2b0f3da5ba43fde12258af6b35c2.ini): failed to open stream: Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/sys/io/File.class.php line 12<br />#0 [internal function]: _hx_error_handler(2, 'file_put_conten...', '/home/config_ad...', 12, Array)
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is there any problem in defining this value?
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What is Schrodinger equation?
What are important features of the quantum mechanical model of atom?
The important features are:
(i) The energy of electrons in atoms is quantized.
(ii) The existence of quantized electronic energy levels is a direct result of the wave like properties of electrons.
(iii) Both the exact position and exact velocity of an electron in an atom can not be determined simultaneously (Heisenberg uncertainty principle).
(iv) An atomic orbital is the wave function ψ for an electron in an atom. Whenever an electron is described by a wave function, we say that the electron occupies that orbital.
(v) The probability of finding an electron at a point within an atom is proportional to [ψ]2 at that point. [ψ]2 is known as probability density and is always positive.
(vi) From the value of [ψ]2 at different points within an atom, it is possible to predict the region around the nucleus where electron will most probably be found.
(vii) In each orbital, the electron has a definite energy. An orbital cannot contain more than two electrons.
(viii) In multi-electron atom, the electrons are filled in various orbitals in order of increasing energy.
ψ is a wave function and refers to the amplitude of electron wave i.e. probability amplitude. It has got no physical significance. The wave function ψ may be positive, negative or imaginary.
[ψ]2 is known as probability density and determines the probability of finding an electron at a point within the atom. This means that if:
(i) is zero, the probability of finding an electron at that point is negligible.
(ii) [ψ]2 is high, the probability of finding an electron is high i.e. electron is present at that place for a long time.
(iii) [ψ]2 is low. the probability of finding an electron is low i.e. electron is present at that place for a shorter time.
In view of Heisenberg’s uncertainty principle, Bohr’s model of the atom in which electrons are treated as particles revolving in definite orbits is meaningless. Thus, its trajectory can not be described as was done by Bohr.
Since in view of Heisenberg’s uncertainty principle, the exact position and momentum of a small moving panicle- like electron can not be determined simultaneously. so the best way is to speak of the probability of finding an electron in a certain region of space around the nucleus rather than describing its specific position. Thus, Heisenberg’s uncertainty principle introduces the concept of probability.
Orbital is the three-dimensional space around the nucleus in which the probability of finding the electron is maximum. In fact, the orbital is the space around a nucleus in which the probability of finding the electron is almost 95%; only for the rest of time, the electron remains out of it.
It is rather difficult to draw the real picture of an orbital. The best way to represent the orbital is in the form of electron density or electron cloud in terms of small dots. The intensity of dots gives the relative probability of finding the electron in that particular region. It may be noted that the probability of finding an electron is never zero even at a large distance from the nucleus. Therefore, an orbital does not have well-defined boundaries.
|
Orbit |
Orbital |
|
1.It is well defined circular path traced by an electron around the nucleus.
|
1. It is the region in space around the nucleus where the probability of finding an electron is maximum.
|
|
2. It represents the planar motion of an electron. |
2. It represents the three-dimensional motion of an electron around the nucleus.
|
|
3. Orbit is circular in shape.
|
3. Orbitals may have different shapes. For example, s-orbital is spherical, p-orbital is dumb-bell shaped and d-orbital is double dumbbell shaped.
|
|
4. The maximum number of electrons in an orbit is 2n- where n is the number of orbits.
|
4. The maximum number of electrons in an orbital is two.
|
|
5. An orbit has a fixed location from the nucleus.
|
5. An orbital is a riot situated at an exact distance from the nucleus.
|
|
6. There is certainty about the position and momentum (or velocity) of an electron in an orbit (Heisenberg's uncertainty principle is violated). |
6. There is no certainty of position and momentum of an electron wave at any point inside the orbital. It obeys uncertainty principle. |
What are quantum numbers? Explain briefly the various quantum numbers which completely specify an electron in an atom.
A set of numbers used to provide a complete description of an electron in an atom is called quantum numbers. There are four quantum numbers required for a complete characterisation of various electrons in an atom. They are:
(i) Principal quantum number (n)
(ii) Azimuthal quantum number (l)
(iii) Magnetic quantum number (m)
(iv) Spin quantum number (s)
(i) Principal quantum number (n): The principal quantum number determines the main energy level or shell to which an electron belongs. It can have all positive integral values such as 1,2, 3,4. etc. The various values of n are also designated by K, L, M, N and so on.
n = 1 2 3 4 5
Orbits = K L M N O
The Principal quantum number gives an idea of
(a) the size of the orbital and the distance of the electron from the nucleus.
(b) the energy of the electron.
As the value of n increases, the electron gets farther away from the nucleus and its energy increases.
The maximum number of electrons in a shell is equal to 2n2 where n is the principal quantum number.
n = 1 2 3 4
Maximum 2x12 2 x 22 2 x 32 2 x 42
electrons (2n2) = 2 = 8 = 18 = 32
(ii) Azimuthal quantum number: Azimutal quantum number:
(a) denotes the sub-shell to which electron belongs,
(b) specifies the angular momentum of the electron and
(c) signifies the shape of the orbital.
It can have all possible whole number values from l=0 to l=n-1, so that the total values of l are equal to the value of principal quantum number, n. Each value of l represents a particular subshell or sub level.
lf n= 1 (K shell) l = 0, (1s subshell) [one value]
n = 2 (L shell) ; l = 0, 1 (2s and 2p subshell respectively) [two values]
n = 3 (M shell) ; l = 0, 1, 2, (3s, 3p and 3d subshell respectively) [three values]
n = 4 (N shell) 1 = 0, 1, 2, 3 (4s, 4p, 4d, 4f subshell, respectively) [four values]
Thus, once n is fixed, lower the value of l, lower will be the energy.
e.g. for M shell (n = 3), 3s < 3p < 3d
4s = 4p = 4d= 4f.
Designation of Azimuthal quantum number (l) by symbols:
|
Value of l |
Designation |
Origin of s p d f |
shape of subshell |
|
l = 0 l= 1 l = 2 l=3 |
s p d f |
sharp principal diffused fundamental |
Spherical Dumb-bell Double dumb-bell Complicated shape |
Maximum number of electrons that can be occupied by a subshell is 2(2l + 1) or 4l + 2.
(iii) Magnetic quantum number (m). An electron due to its angular motion around the nucleus produces a magnetic field which can interact with the external magnetic field. Thus, under the influence of the external magnetic field, the electrons in a given sub-shell orient themselves in certain specific regions of space around the nucleus. These regions of space are called orbitals. This quantum number determines the number of the preferred orientations of electrons in a subshell in space around the nucleus. For a given value of l, m can have values from - l to + l including zero (-l to zero to +l) and in all (2l + 1) values. For example.
If l = 0 ; m = 0
l = I ; m = - 1, 0 + 1
l = 2 ; m -2,- 1,0 ,+1, +2
l = 3 ;m =-3,-2,- 1,0, +1, +2, +3
It means that s-subshell (l = 0) has one orientation of electron i.e. s orbital:
p-subshell (l = 1) has three orbitalspx,py,pz d-subshell (l = 2) has five orientations i.e. has five orbitals,
f-subshell (l = 3) has seven orientations of electrons i.e. f-subshell has seven orbitals.
(iv) Spin quantum number (s). Electron revolves around the nucleus and at the same time it rotates about its own axis. An electron may have two possible spins -clockwise or anticlockwise. Therefore quantum s can have one of the two possible values +1/2and -1/2 depending upon the direction of the spin. Instead of numbers + 1/2 and -1/2 , the directions of spin are designated by arrows pointing up and down (↑ or ↓) respectively.
Significance of quantum numbers:
(i) Principal quantum number (n): It signifies the size of the electron cloud. It can have values 1, 2, 3, 4.....etc.
(ii) Azimuthal quantum number (l): It signifies the shape of electron cloud. It can have values from l = 0 to l = n - 1.
(iii) Magnetic quantum number (m): It signifies the orientation of electron cloud. It can have values from - l to + l including zero.
(iv) Spin quantum number (s): It signifies the spin of the electrons can have values ![<pre>uncaught exception: <b>file_put_contents(/home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/web/../../../../../../formulas/88/8f/343ad109b0f5e52b813049296525.ini): failed to open stream: Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/sys/io/File.class.php at line #12file_put_contents(/home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/web/../../../../../../formulas/88/8f/343ad109b0f5e52b813049296525.ini): failed to open stream: Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/sys/io/File.class.php line 12<br />#0 [internal function]: _hx_error_handler(2, 'file_put_conten...', '/home/config_ad...', 12, Array)
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1/2 for each value of m.
An electron is placed in 4f orbital. What possible values for the quantum numbers, n, l, m and s can it have?
For 4f orbital, n = 4, l = 3
When l=3, m = -3, -2. -1, 0 + 1, +2, +3. (seven values)
For each value of m, there are two possible values for s i.e. + 1/2 and - 1/2 .
(i) An atomic orbital has n = 3. what are the possible values of l and ml ?
(ii) List the quantum number (ml and l) of electrons for 3d orbital.
(iii) Which of the following orbitals are possible: 1p, 2s, 2p, and 3f?
(i) For n = 3; l = 0, 1 and 2
For l = 0, ml = 0
For l =1, ml = -1, 0+1
For l = 2, ml = -2, -1, 0, +1, +2
(ii) For 3d orbital, n = 3, l =2
For l =2, ml = -2, -1, 0, +1, +2
(iii) 1 p is not possible because when n = 1, l = 0 only (For p, l = 1)
2s is possible because when n =2, l = 0, 1(For s, l = 0)
2p is possible because when n=2, l = 0, (For p, l = 1)
3f is not possible because when n = 3, l = 0, 1, 2(For f, l = 3).
An atomic orbital has n = 3, What are the possible value of l?
When n =3, l= 0,1,2.
When l=0, ml =0
When l=1, ml =-1,0, +1.
When l=2, m2 = -2, -1, 0, +1, +2.
Give all the possible values of l, m and s for electrons when n = 3.
When n = 3 ; l= 0, 1, 2.
l = 0. l = 1, l = 2
For l = 0, m = 0
For l= 1, m = -1, 0, + 1
For l = 2, m = -2,-1,0, +1, +2
For each value of m, there are two possible values of s i.e. + 1/2 and -1/2 .
Using s, p, d notations describe orbital with the following quantum numbers:
(a) n = 1, l = 0; (b) n = 3; l = 1 (c) n = 4; l = 2; (d) n =4, l = 3.
Orbital of the given quantum number,
(a) 1s
(b) 3p
(c) 4d
(d) 4f
(i) Not possible because n cannot be 0.
(ii) Possible.
(iii) Not possible as when n= 1, l cannot be 1.
The number of orbitals in 4th shell = n2 - 42 = 16 Since each orbital can accommodate a maximum of two electrons with opposite spin, therefore each orbital has one electron with ms = - 1/2 .
Hence there will be 16 electrons with ms = - 1/2 .
The given set of quantum number is not possible because the value of the principal quantum number (n) cannot be zero.
From the following sets of quantum numbers, state which are possible. Explain why others are not permitted.
(i) n = 1, l = 0, m = +1, s = +1/2
(ii) n = 2, l = 1, m =-1, s = -1/2
(iii) n = 2, l =2, m = 0, s = -1/2
(iv) n =2, l =1, m = 0, s =+1/2
(i) The given sets of quantum numbers are not possible ; when l = 0, m cannot be + 1.
(ii) The given sets of quantum numbers are Possible.
(iii) The given sets of quantum numbers are not possible because when n = 2, l cannot be 2.
(iv) The given sets of quantum numbers are Possible.
What subshells are possible in n = 3 energy level?
For n = 3 energy level, the possible values of l are 0, 1 and 2. Thus, there are three subshells namely:
l = 0, s subshell ; l = 1, p subshell ; and l = 2, d subshell.
For g- orbital, l=4,
As for any value ‘n’ of principal quantum number, the Azimuthal quantum number (l) can have a value from zero to (n-1).
Therefore, for l=4, a minimum value of n=5.
State and explain Pauli's Exclusion Principle.
Pauli’s Exclusion Principle: It states: No two electrons in an atom can have the same set of all the four quantum numbers (n, l, m, s).
According to this principle, two electrons in an atom can have a maximum of three quantum numbers identical but the value of the fourth quantum number must be different. Electrons having the same set of values n, l and m are said to belong to the same orbital.
For example, for K shell i.e. when n = 1
Thus, we conclude from the above that:
(i) an orbital cannot have more than 2 electrons, (ii) if an orbital has two electrons, they must have opposite spin.
Application of Pauli’s Exclusion Principle: The maximum number of electrons in different sub-shells and energy levels can be deduced from this principle.
For L shell (n = 2)
|
n |
l |
m |
s |
|
2 2 2 2 2 2 2 2 |
0 0 1 1 1 1 1 1 |
0 0 -1 -1 0 0 +1 +1 |
+1/2 -1/2 +1/2 +1/2 -1/2 -1/2 +1/2 -1/2 |
Discuss the shapes of s-orbitals.
| 1s orbital | 2s orbital |
| 1. It is smaller in size than 2s orbital. | 1. It is larger in size than 1s orbital. |
| 2. It consists of a single sphere having no local plane. | 2. It consists of two spheres having one nodal plane. |
| 3. An electron in it has lesser energy. | 3. An electron in it has greater energy. |
The bromine atom possesses 35 electrons. It contains 6 electrons in 2p orbital, 6 electrons in 3p orbital and 5 electrons in 4p orbital. Which of these electron experiences the lowest effective nuclear charge?
Force of attraction between nucleus and the electron depend upon the distance between the nucleus and the orbital, in which electron present. As the distance increases, the effective nuclear charge also decreases. Since 4p electrons being farthest as compared to 2p and 3p electrons from the nucleus, therefore 4p electrons will experience the lowest effective nuclear charge.
Among the following pairs of orbitals which orbital will experience the large effective nuclear charge? (i) 2s and 3s (ii) 4d and 4f (iii) 3d and 3p.
Force of attraction between nucleus and the electron present around it depend upon the distance between the nucleus and the orbital, in which electron present. As the distance increases, the effective nuclear charge also decreases.
(i) Out of 2s and 3s orbitals: 2s orbital is closer to the nucleus than 3s orbital. Hence 2s orbital will experience larger effective nuclear charge.
(ii) Out of 4d and 4f orbitals: 4d orbital is closer to the nucleus than 4f orbital. Hence 4d orbital will experience larger effective nuclear charge.
(iii) Out of 3d and 3p orbitals: 3p orbital is closer to the nucleus than 3d orbital. Hence 3p orbital will experience large effective nuclear charge.
The unpaired electrons in Al and Si are present in 3p orbital. Which electrons will experience more effective nuclear charge from the nucleus ?
Discuss the shapes of p-orbitals.
How will you differentiate between s- and p-orbitals?
The difference between s- and p-orbitals:
| s-orbtial | p-orbital |
| 1. It is spherical in shape. | 1. It is dumb-bell in shape. |
| 2. It is non-directional in nature. | 2. It is directional in nature. |
| 3. The value of l = 0 | 3. The value of l = 1 |
4. It cannot participate in  bond formation. |
4. It can take part in bond formation. |
What is Hund's rule of Maximum Multiplicity?
For example, each of the three p-orbitals of the p-subshell, each of the five d-orbitals of d-subshell and each of the seven f-orbitals of the f-subshell gets one electron of parallel spin before any other of them receives the second electron of opposite spin. According to Hund’s rule, the electronic configuration of nitrogen (7N) is
When there is a group of empty orbitals with equal energies (2px, 2py, 2pz), why are electrons first alloted singly to different orbitals ?
Electronic configuration of first ten element:
|
Element |
Atomic number (z) |
Electronic configuration |
|
1.Hydrogen (H) |
1 |
1s1 |
|
2.Helium (He) |
2 |
1s2 |
|
3.Lithium (Li) |
3 |
1s2 2s1 |
|
4.Berylium (Be) |
4 |
1s2 2s2 |
|
5.Boron (B) |
5 |
1s2 2s2 2p1 |
|
6. carbon (C) |
6 |
1s2 2s2 2p2 |
|
7.Nitrogen (N) |
7 |
1s2 2s2 2p3 |
|
8.Oxygen (O) |
8 |
1s2 2s2 2p4 |
|
9.Fluorine (F) |
9 |
1s2 2s2 2p5 |
|
10. Neon (Ne) |
10 |
1s2 2s2 2p6 |
Write the electronic configuration of elements from atomic number 11 to atomic number 20. Also name the elements.
Electronic configuration of the element from atomic number 11 to 20.
|
Element |
Atomic number (Z) |
Electronic configuration |
|
Sodium (Na) |
11 |
[Ne]10 3s1 |
|
Magnesium (Mg) |
12 |
[Ne]10 3s2 |
|
Aluminium (Al) |
13 |
[Ne]10 3s2 3p1 |
|
Silicon (Si) |
14 |
[Ne]10 3s2 3p2 |
|
Phosphorus (P) |
15 |
[Ne]10 3s2 3p3 |
|
Sulphur (S) |
16 |
[Ne]10 3s2 3p4 |
|
Chlorine (Cl) |
17 |
[Ne]10 3s2 3p5 |
|
Argon (Ar) |
18 |
[Ne]10 3s2 3p6 |
|
Potassium (K) |
19 |
[Ar]10 4s1 |
|
Calcium (Ca) |
20 |
[Ar]10 4s2 |
Using the Aufbau principle, write the electronic configuration for the ground state of the following atoms:
(i) Boron (Z = 5)
(ii) Neon (Z = 10)
(iii) Al (Z = 13)
(iv) Chlorine (Z = 17)
(v) Calcium (Z = 20).
(i) Boron (Z = 5),
Electronic configuration = 
(ii) Neon (Z = 10),
Electronic configuration = 
(iii) Al(Z =13)
Electronic configuration
(iv) Chlorine (Z = 17),
Electronic configuration 
(v) Calcium (Z = 20),
Electronic configuration
Give the electronic configuration of following ions:
(i) H- (ii) F- (iii) Mg2+ (iv) S2-
(i) H- (Z = 1)
Electronic configuration = 1s2
(ii) F- (Z = 9),
Electronic configuration = ![<pre>uncaught exception: <b>file_put_contents(/home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/web/../../../../../../formulas/9e/9b/e07a1bb8bbc5c38e1b6830e2c075.ini): failed to open stream: Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/sys/io/File.class.php at line #12file_put_contents(/home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/web/../../../../../../formulas/9e/9b/e07a1bb8bbc5c38e1b6830e2c075.ini): failed to open stream: Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/sys/io/File.class.php line 12<br />#0 [internal function]: _hx_error_handler(2, 'file_put_conten...', '/home/config_ad...', 12, Array)
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#7 {main}</pre>](/application/zrc/images/qvar/CHEN11085554.png)
(iii) Mg2+ (Z = 12),
Electronic configuration = 
(iv) S2- (Z = 16);
Electronic configuration![<pre>uncaught exception: <b>file_put_contents(/home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/web/../../../../../../formulas/eb/8d/802d3b5341981d70d750f920e034.ini): failed to open stream: Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/sys/io/File.class.php at line #12file_put_contents(/home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/web/../../../../../../formulas/eb/8d/802d3b5341981d70d750f920e034.ini): failed to open stream: Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/sys/io/File.class.php line 12<br />#0 [internal function]: _hx_error_handler(2, 'file_put_conten...', '/home/config_ad...', 12, Array)
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#7 {main}</pre>](/application/zrc/images/qvar/CHEN11085554-2.png)
Write down the electronic configuration of elements from atomic number 21 to atomic number 30. Also name the elements.
| Element | Atomic number (Z) | Electronic Configuration |
| Scandium (Sc) | 21 | [Ar]184s23d1 |
| Titanium (Ti) | 22 | [Ar]18 4s23d2 |
| Vanadium(V) | 23 | [Ar]18 4s23d3 |
| Chromium(Cr) | 24 | [Ar]18 4s23d5 |
| Manganese(Mn) | 25 | [Ar]18 4s23d5 |
| Iron (Fe) | 26 | [Ar]18 4s23d6 |
| Cobalt (Co) | 27 | [Ar]18 4s23d7 |
| Nickel (Ni) | 28 | [Ar]18 4s23d8 |
| Copper (Cu) | 29 | [Ar]18 4s13d10 |
| Zinc (Zn) | 30 | [Ar]184s23d10 |
How will you account for the electronic configuration of the elements with atomic number of 24 and 29?
Or
Account for the following:
The continuous building up of 3d subshell interpreted in chromium (At. No. 24) and copper (At No. 29).
Or
Outer electronic configuration of chromium is 3d5 4s1 and not 3d4 4s2. Explain.
Atomic number (24):![<pre>uncaught exception: <b>file_put_contents(/home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/web/../../../../../../formulas/67/3e/c742081b39dbe7d8ca8d0fb6e80d.ini): failed to open stream: Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/sys/io/File.class.php at line #12file_put_contents(/home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/web/../../../../../../formulas/67/3e/c742081b39dbe7d8ca8d0fb6e80d.ini): failed to open stream: Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/sys/io/File.class.php line 12<br />#0 [internal function]: _hx_error_handler(2, 'file_put_conten...', '/home/config_ad...', 12, Array)
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#7 {main}</pre>](/application/zrc/images/qvar/CHEN11085562.png)
The above configuration is less symmetrical and also less stable because four 3d orbtials are half killed while one is empty. But the actual electronic configuration of chromium is:
Since all the five 3d orbitals are half filled so this configuration is more symmetrical and also more stable.
(ii) Atomic number (29)
But the actual electronic configuration of copper is
i.e. all the 3d orbitals get completely filled and 4s orbital is exactly half filled. Hence this configuration is more symmetrical and also more stable.
Write the electronic configurations of the following ions: (a)H- (b) Na+ (c) O2- (d) F-
(a) H- (Z = 1);
Electronic configuration = 1s2
(b) Na+ (Z = 11);
Electronic configuration = 1s22s22p6
(c) O2-(Z = 8);
Electronic configuration = 1s22s22p6
(d) F-(Z = 9);
Electronic configuration = 1s22s22p6
(a) 1s22s22p63s1;
Atomic number of the element is 11.
(b) 1s22s22p3;
Atomic number of the element is 7.
(c) 1s22s22p63s23p5;
Atomic number of the element is 17.
Which atoms are indicated by the following configurations?
(a) [He] 2s1
(b) [Ne] 3s2 3p3
(c) [Ar] 4s2 3d1.
(a) [He] 2s1, the atom is lithium (3Li).
(b) [Ne] 3s2 3p3, the atom is phosphours (15P).
(c) [Ar] 4s2 3d1, the atom is scandium ( 21Sc).
Which of the following are isoelectronic species, i.e. those having the same number of electrons?
Na+, K+ , Mg2+, Ca2+, S2-, Ar.
Hence isoelectronic species are
(i) Na+ and Mg2+
(having number of electrons = 10).
(ii) K+, Ca2+, S2- and Ar (having number of electrons = 18).
What is the cause of greater stability of exactly half-filled and completely filled configurations?
The greater stability of these configurations is due to following two reasons:
(i) Symmetry. The half-filled and completely filled configurations are more symmetrical and symmetry leads to greater stability.
(ii) Exchange energy. In an atom, the electrons present in various orbitals of the same sub-shell tend to exchange their positions. Such an exchange of electrons results in the release of energy called exchange energy. Greater the exchange energy, more is the stability associated with the orbitals. Now half filled and completely filled sub-shell configurations allow the maximum exchange of electron to take place. Therefore, such configurations have maximum stability.
An atom of an element contains 29 electrons and 35 Deutrons. Deduce (i) the number of protons and (ii) the electronic configuration of the element.
We know for neutral atom,
Number of protons = Number of electrons = 29
Thus atomic number of the element = 29
Electronic configuration of element (Z = 29) can be represented as 29Z ;
1s22s22p63s23p64s13d10
or [Ar]184s13d10
Write the electronic configurations of the following ions/atom:
(i) Cu2+ (ii) Cr3+
(iii) Ni2+ (iv) Rubidium (Z = 37)
The electronic configuration of the given atom/ion is,
(i)Cu2+ :
Electronic configuration of Cu2+ (Z = 29) is 1s2 2s2 2p6 3s2 3p6 3d9
[∵ number of electrons = 29 - 2 = 27]
(ii) Cr3+:
Electronic configuration of Cr3+ (Z = 24) is 1s2 2s2 2p6 3s2 3p6 3d3
[∵ number of electrons = 24 - 3 = 21]
(iii) Ni2+:
Electronic configuration of Ni2+ (Z = 28) is 1s2 2s2 2p6 3s2 3p6 3d8
[∵ number of electrons = 28 - 2 = 26]
(iv) Rubidium (Z = 37):
Electronic configuration of Rb (Z = 37) is
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s1
Write the value of n, l and m for the electron with the highest energy in sodium atom (Z =11).
The electronic configuration of Na
(Z= 11) is 1s2 2s22p6 3s1
For 3s electron
n = 3; L = 0, m = 0.
With the help of ground state electronic configuration of element Cr(Z= 24), deduce the possible values of four quantum numbers for the 19th electron of Cr.
Here T = 2.0 x 10-10 s![<pre>uncaught exception: <b>file_put_contents(/home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/web/../../../../../../formulas/f3/70/d1d4a51e5ebb15c78a2cabd2dd2e.ini): failed to open stream: Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/sys/io/File.class.php at line #12file_put_contents(/home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/web/../../../../../../formulas/f3/70/d1d4a51e5ebb15c78a2cabd2dd2e.ini): failed to open stream: Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/sys/io/File.class.php line 12<br />#0 [internal function]: _hx_error_handler(2, 'file_put_conten...', '/home/config_ad...', 12, Array)
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#7 {main}</pre>](/application/zrc/images/qvar/CHEN11085705.png)
Diameter of carbon atom = 0.15 nm
= 0.15 x 10-9 m = 1.5 x 10-10 m
Length along which carbon atoms are to be placed
= 20 cm
= 20 x 10-2m = 2 x 10-1 m
Hence, number of carbon atoms which can be placed along the length
A stream of electrons from a heated filament was passed between two charged plates kept at a potential difference V esu. If e and m are charge and mass of an electron, respectively, then the value of h/λ (where λ is wavelength associated with electron wave) is given by:
2meV
mev
C.
The relation between h/λ and energy is given as:
Applying de-Broglie wavelength and kinetic energy term in eV.
de-Broglie wavelength for an electron (λ) = h/p
⇒ p = h/ λ (i)
Kinetic energy of an electron = eV
As we know that,
From equations (i) and (ii), we get
Which of the following is the energy of a possible excited state of hydrogen?
+13.6 eV
-6.8 eV
-3.4 eV
+6.8 eV
C.
-3.4 eV
Since at n=1 the population of electrons is maximum i.e. at the ground state. So, maximum excitation will take place from n = 1 to n=2.
Hence, n=2 is the possible excited state,
Now, we have the formula for energy of H-atom
(En)H = 
where Z = atomic number
Z for H-atom = 1
therefore, (En)H = 
The correct set of four quantum numbers for the valence electrons fo rubidium atom (Z = 37) is
5,0,0, +1/2
5,1,0,6+1/2
5,1,1,+1/2
5,0,1,+1/2
A.
5,0,0, +1/2
the atomic number of Rb (Z) = 37
Thus, its electronic configuration [Kr]5s1. since the last electron or valence electron enters in 5s subshell. So, the quantum numbers are n = 5, l =0(for s orbital) m=0
(m = +l to -l), s = +1/2 or -1/2
The electrons identified by quantum numbers n and l:
(a) n = 4, l = 1 (b) n = 4, l = 0 (c) n = 3, l = 2 (d) n = 3 , l = 1
Can be placed in order of increasing energy as
(c) < (d) < (b) < (a)
(d) < (b) < (c) < (a)
(b) < (d) < (a) < (c)
(a) < (c) < (b) < (d)
B.
(d) < (b) < (c) < (a)
a.) (n + l) = 4 + 1 = 5
b.) (n + l) = 4 + 0 = 4
c.) (n + 1) = 3 + 2 = 5
d.) (n + 1) = 3 + 1 = 4
Higher the value of (n+1), higher the energy. if (n+1) are same, sub orbit with a lower value of n has lower energy. Thus,
3p<4s<3d<4p
The ionization energy of He+ is 19.6 x 10–18 J atom–1. The energy of the first stationary state (n = 1) of Li2+ is
4.41 x 10-16 J atom–1
-4.41 x 10-17 J atom–1
-2.2 x 10-15 J atom–1.
8.82 x 10-17 J atom–1.
B.
-4.41 x 10-17 J atom–1
Ionisation energy of He+
= 19.6 × 10-18 J
E1 (for H) × Z2= IE
E1 × 4 = – 19.6 × 10-18 J
E1 (for Li2+) = E1 for H × 9
-4.41 x 10-17 J atom–1
The radius of the second Bohr orbit for the hydrogen atom is :
(Plank's const. h = 6.6262 × 10–34 Js ; mass of electron = 9.1091 × 10–31 kg ; charge of electron
e = 1.60210 × 10–19 C ; permittivity of vacuum
∈0 = 8.854185 × 10–12 kg–1 m–3 A2)
1.65Å
4.76Å
0.529Å
2.12Å
D.
2.12Å
Radius of nth Bohr orbit in H-atom = 0.53 n2Å
Radius of II Bohr orbit = 0.53 ×(2)2 = 2.12 Å
Which of the following sets of quantum numbers is correct for an electron in 4f orbital?
n = 4, I =3, m = +4, s = + 1/ 2
n = 3, I = 2, m = -2, S = + 1/2
n =4, I = 3, m = +1, s = + 1/ 2
n =4, I = 4, m -4, s = -1/ 2
C.
n =4, I = 3, m = +1, s = + 1/ 2
For 4f orbital electrons, n = 4
s p d f
l = 3 (because 0 1 2 3 )
m = +3, +2, +1, 0, –1, –2, –3
s = ±1/2.
Two electrons occupying the same orbital are distinguished by
Magnetic quantum number
Azimuthal quantum number
Spin quantum number
Principal quantum number
C.
Spin quantum number
Two electrons occupying the same orbital has equal spin but the directions of their spin are opposite. Hence, spin quantum number, s(represented +1/2 and -1/2) distinguishes them.
Magnetic moment 2.84 BM is given by (At. no. Ni= 28, Ti= 22, Cr=24, Co = 27)
Ni2+
Ti3+
Cr3+
Co2+
A.
Ni2+
The value of Planck's constant is 6.63 x 10-34 Js. The speed of light is 3 x1017 nms-1 . Which value is closet to the wavelength in nanometer of a quantum of light with frequency of 6 x 1015 s-1 ?
10
25
50
75
C.
50
Given, Planck's constant,
h= 6.63 x10-34
speed of light, c= 3 x1017 nms-1
Frequency of quanta
v=6 x1015 s-1
Wavelength, λ =?
We know that,
What is the maximum number of electrons that can be associated with the following set of quantum number?
n=3, l =1 and m=-1.
10
6
4
2
D.
2
The orbital of the electron having =3, l =1 and m= -1 is 3pz (as nlm) and an orbital can have a maximum of two electrons with opposite spins.
therefore, 3pz orbital contains only two electrons or only 2 electrons are associated with n=3, l=1, m=-1.
Based on equation
E=-2.178 x 10-18 J 
certain conclusions are written. Which of them is not correct?
The negative sign in the equation simply means that the energy of an electron bound to the nucleus is lower than if would be if the electrons were at the infinite distance from the nucleus.
Larger the value of n, the larger is the orbit radius
Equation can be used to calculate the change in energy when the electron changes orbit
For n=1 the electron has a more negative energy than it does for n=6 which means that the electron is more loosely bound in the smallest allowed orbit.
D.
For n=1 the electron has a more negative energy than it does for n=6 which means that the electron is more loosely bound in the smallest allowed orbit.
If n=1,
E1 = - 2.178 x 10-18 Z2 J
If n=6
From the above calculation, it is obvious that electron has a more negative energy than it does for n=6. it means that electron is more strongly bound in the smallest allowed orbit.
Maximum number of electrons in a subshell with l =3 and n=4 is
14
16
10
12
A.
14
n represents the main energy level and l represents the subshell.
If n=4 and l = 3, the subshell is 4f.
If f subshell, there are 7 orbitals and each orbital can accommodate a maximum number of electrons, so, maximum number of electrons in 4f subshell = 7 x 2 = 14
The correct set four quantum number for the valence electron of rubidium atom (z=37) is
5,1,1, +1/2
6,0,0,+1/2
5,0,0 +1/2
5,1,0, +1/2
C.
5,0,0 +1/2
37Rb = 36[Kr] 5s1The total number of atomic orbitals in fourth energy level of an atom is
16
32
4
8
A.
16
Number of atomic orbitals in an orbit is given by,
= n2 = 42 = 16
The number of atomic orbitals in fourth energy level is = 16
The energies E1 and E2 of two radiations are 25 eV respectively. The relation between their wavelength i.e., λ1 and λ2 will be
λ1 = 2λ2
λ1 = 4λ2
λ1 = λ2/2
λ1 = λ2
A.
λ1 = 2λ2
E1 = 25eV. E2 = 50 eV
E1 = hc/λ1 and E2 = hc/λ2
If n =6, the correct sequence for filling of electrons will be
ns → (n-1) d → ( n - 2)f → np
ns - (n-2)f → np → (n-1)d
ns - np → (n-1)d → (n-2)f
ns → (n-2) f → (n-1)d →np
D.
ns → (n-2) f → (n-1)d →np
6s → 4f → 5d → 6p for n = 6
Which is the correct order of increasing energy of the listed orbitals in the atom of titanium?
3s 4s 3p 3d
4s 3s 3p 3d
3s 3p 3d 4s
3s 3p 4s 3d
C.
3s 3p 3d 4s
According to Aufbau rule
3s < 3p < 3d< 4s
According to the Bohr theory, which of the following transitions in the hydrogen atom will give rise to the least energetic photon?
n= 6 to n = 1
n = 5 to n = 4
n= 6 to n = 5
n = 5 to n = 3
C.
n= 6 to n = 5
Maximum number of electrons in a subshell of an atom is determined by the following
4l + 2
2l +1
4l-2
2n2
A.
4l + 2
Total number of subshells = (2l +1)
therefore,
Maximum number of electrons in the subshell = 2 (2l +1) = 4l + 2
The energy absorbed by each molecule (A2) of a substance is 4.4 x 10-19 J and bond energy per molecule is 4.0 X 10-19 J. The kinetic energy of the molecule per atom will be
2.0 x 10-20 J
2.2 x 10-19 J
2.0 x 10-19 J
4.0 x 10-20 J
A.
2.0 x 10-20 J
KE of molecule = energy absorbed by molecules - BE per molecule
= [4.4 x 10-19)-(4.0 x 10-19) J
= 0.4 x 10-19 J
KE per atom 0.4 x 10-19/2 J = 2.0 x 10-20 J
Which of the following is not permissible arrangement of electrons in an atom?
n = 4, l=0, m=0, s=-1/2
n=5, l=3, m=0, s=+1/2
n=3, l=2, m=-3, s= -1/2
n=3, l=2, m=-2, s=-1/2
C.
n=3, l=2, m=-3, s= -1/2
For an electron, n may be 0,1, 2....and l=0 to n=-1 and m = - l to l (including 0) and 
1/2.
Hence, If ,
n=3,
l=0 to (3-1)
=0,1,2
m=-l to+l
=-2,-1,0, +1,+2
s 
1/2
If uncertanity in position and momentum are equal, then uncertaninty in velocity is
A.
According to Heisenberg's uncertainty principle, it is impossible to determine simultaneously the position and momentum of moving a particle, ie.
The uncertainty principle in terms of energy and time is given as
Given: The mass of electron is 9.11 x 10-31 kg
Planck constant is 6.626 x 10-34 Js,
the uncertainty involved in the measurement of velocity within a distance of 0.1 A is:
5.79 x 106 ms-1
5.79 x 107 ms-1
5.79 x 108 ms-1
5.79 x 105 ms-1
A.
5.79 x 106 ms-1
By Heisenberg's uncertainty principle
The orientation of an atomic orbital is governed by:
azimuthal quantum number
spin quantum number
magnetic quantum number
the principal quantum number
C.
magnetic quantum number
The orientation of an atomic orbital is governed by magnetic quantum number.
Which one is the wrong statement?
de-Broglie's wavelength is given by λ = h/mv,
where m = mass of the particle, v = group velocity of the particle
The uncertainty principle is 
Half-filled and fully filled orbitals have greater stability due to greater exchange energy, greater symmetry and more balanced arrangement
The energy of 2s orbital is less than the energy of 2p orbital in case of Hydrogen like atoms
D.
The energy of 2s orbital is less than the energy of 2p orbital in case of Hydrogen like atoms
Which one is a wrong statement?
Total orbital angular momentum of electron in 's' orbital is equal to zero
An orbital is designated by three quantum numbers while an electron in an atom is designated by four quantum numbers
The value of m for dz2 is zero
The electronic configuration of N atom is
D.
The electronic configuration of N atom is
According to Hund's Rule of maximum multiplicity, the correct electronic configuration of N-atom is
Magnesium reacts with an element (X) to form an ionic compound. If the ground state electronic configuration of (X) is 1s2 2s2 2p3, the simplest formula for this compound is
Mg2X3
MgX2
Mg3X2
Mg2X
C.
Mg3X2
Element (X) electronic configuration
1s2 2s2 2p3
So, valency of X will be 3.
Valency of Mg is 2.
Formula of the compound formed by Mg and X
will be Mg3X2
Among the following set of quantum numbers, the impossible set is
| n | l | m | s |
| 3 | 2 | -3 | -1/2 |
| n | l | m | s |
| 4 | 0 | 0 | 1/2 |
| n | l | m | s |
| 5 | 3 | 0 | -1/2 |
| n | l | m | s |
| 3 | 2 | -2 | 1/2 |
A.
| n | l | m | s |
| 3 | 2 | -3 | -1/2 |
Principal quantum number = n
Azimuthal quantum number = l = 0 to (n-1)
Magnetic quantum number = m = -l to +1
Spin quantum number = s = +1/2 or -1/2
Now,
(i) In first option the given values are,
n=3; l=2; m=-3; s=-1/2
So according to Azimuthal quantum number
l = 0 to (n-1) = 0 to 2 = here it is 2 which is permissible
m = -l to +l = -2 to +2 = here it is 3 which is not permissible
s = +1/2 or -1/2 = here it is -1/2 whch is permissible
(ii) In the second option given values are,
n = 4; l = 0, m = 0; s= 1/2
So according to Azimuthal quantum number
l = 0 to (n-1) = 0 to 3 = here it is 0 which is permissible
m =-l to +l = -3 to +3 = here it is 0 which is permissible
s = +1/2 or -1/2 = here it is 1/2 whch is permissible
(iii) In the third option given values are,
n =5; l = 3; m= 0; s= -1/2
So according to Azimuthal quantum number
l = 0 to (n-1) = 0 to 4 = here it is 3 which is permissible
m = -l to +l = -4 to +4 = here it is 0 which is permissible
s = +1/2 or -1/2 = here it is -1/2 whch is permissible
(iv) it the forth option given values are,
n=3; l =2; m=-2; s = 1/2
So according to Azimuthal quantum number
l = 0 to (n-1) = 0 to 2 = here it is 2 which is permissible
m = -l to +l = -2 to +2 = here it is 2 which is permissible
s = +1/2 or -1/2 = here it is -1/2 whch is permissible
Match the type of series given in Column I with the wavelength range given in Column II and choose the correct option.
| Column I | Column II | ||
| A | Lyman | 1. | Ultraviolet |
| B. | Paschen | 2. | Near-infrared |
| C. | Balmer | 3. | Far Infrared |
| D. | Pfund | 4. | Visible |
| A | B | C | D |
| 1 | 2 | 4 | 3 |
| A | B | C | D |
| 4 | 3 | 1 | 2 |
| A | B | C | D |
| 3 | 1 | 2 | 4 |
| A | B | C | D |
| 4 | 3 | 2 | 1 |
A.
| A | B | C | D |
| 1 | 2 | 4 | 3 |
The correct match is
| A → Lyman ultraviolet → (1) Ultraviolet |
| B → Paschen → (2) → Near Infrared |
| C → Balmer → (4) → Visible |
| D → Pfund → (3) → Far Infrared |
The electrons identified by quantum numbers n and l, are as follows
n = 4, l = 0
n =3, l= 2
If we arrange them in order of increasing energy, i.e., from lowest to highest, the correct order is
IV<II<III<I
II<IV<I<III
I < III < II < IV
III < I < IV < II
A.
IV<II<III<I
Reason:
(i) More be the sum of n + l → more be the energy.
(ii) For the
same sum, the more value of n, have higher energy.
Because, n + l for →
(a) 4 + 1 = 5
(b) 4 + 0 = 4
(c) 3+ 2 = 5
(d) 3+1 = 4
therefore, correct order is (iv)<(ii) < (iii) < (i)
Sponsor Area
Sponsor Area
.
is neutral
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#6 [internal function]: _hx_lambda->execute('Http Error #404')
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#8 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/haxe/Http.class.php(444): com_wiris_plugin_impl_HttpImpl->onError('Http Error #404')
#9 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/haxe/Http.class.php(458): haxe_Http->customRequest(true, Object(haxe_io_BytesOutput), NULL, NULL)
#10 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php(40): haxe_Http->request(true)
#11 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(80): com_wiris_plugin_impl_HttpImpl->request(true)
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/php/Boot.class.php(769): call_user_func_array('com_wiris_plugi...', Array)
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#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/haxe/Http.class.php(532): call_user_func_array(Array, Array)
#4 [internal function]: haxe_Http_5(true, Object(com_wiris_plugin_impl_HttpImpl), Object(com_wiris_plugin_impl_HttpImpl), Array, Object(haxe_io_BytesOutput), true, 'Http Error #404')
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/php/Boot.class.php(769): call_user_func_array('haxe_Http_5', Array)
#6 [internal function]: _hx_lambda->execute('Http Error #404')
#7 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php(27): call_user_func_array(Array, Array)
#8 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/haxe/Http.class.php(444): com_wiris_plugin_impl_HttpImpl->onError('Http Error #404')
#9 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/haxe/Http.class.php(458): haxe_Http->customRequest(true, Object(haxe_io_BytesOutput), NULL, NULL)
#10 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php(40): haxe_Http->request(true)
#11 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(80): com_wiris_plugin_impl_HttpImpl->request(true)
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#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
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#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(48): sys_io_File::saveContent('/home/config_ad...', 'mml=<math xmlns...')
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(112): com_wiris_util_sys_Store->write('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
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#7 {main}</pre>](/application/zrc/images/qvar/CHEN11085727.png)
