Structure of Atom

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Question
CBSEENCH11008373

A 0.66 kg ball is moving with a speed of 100 m/s. The associated wavelength will be
(h = 6.6 x 10-34 Js)

  • 6.6 x 10-32 m

  • 6.6 x 10-34 m

  • 1 x 10-35 m

  • 1 x 10-32

Solution

C.

1 x 10-35 m

straight lambda space equals space straight h over mv space equals space fraction numerator 6.6 space straight x space 10 to the power of negative 34 end exponent over denominator 0.66 space straight x space 100 end fraction space equals space 10 to the power of negative 35 end exponent

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Question
CBSEENCH11008338

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 

Solution

C.

n= 6 to n = 5 

increment straight E space proportional to open square brackets space fraction numerator 1 over denominator straight n subscript 1 superscript 2 end fraction minus fraction numerator 1 over denominator straight n subscript 2 superscript 2 end fraction close square brackets space comma space where space straight n subscript 2 greater than straight n subscript 1
Energy of photon obtained from the transition n = 6 to n = 5 will have least energy.

Question
CBSEENCH11008255

Based on equation
E=-2.178 x 10-18open parentheses straight Z squared over straight n squared close parentheses 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.

Solution

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
space straight E subscript 6 end subscript space equals space fraction numerator negative 2.178 space straight x space 10 to the power of negative 18 end exponent over denominator 36 end fraction

equals space 6.05 space straight x space 10 to the power of negative 20 end exponent space straight Z squared space straight J
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.

Question
CBSEENCH12011010

Calculate the energy in joule corresponding to light of wavelength 45mm: (Planck's constant h=6.63 x10-34)Js; speed of light c= 3 x 108 ms-1)

  • 6.67 x 1015

  • 6.67 x 1011

  • 4.42 x 10-15

  • 4.42 x 10-18

Solution

D.

4.42 x 10-18

The wavelength of light is related to its energy by the equation,
straight E space equals hc over straight lambda
Given comma
straight lambda space equals space 45 space mm space equals space 45 space straight x space 10 to the power of negative 9 end exponent space straight m space
left square bracket therefore space 1 space nm space equals space 10 to the power of negative 9 end exponent space straight m space right square bracket

Hence comma space straight E equals fraction numerator 6.63 space straight x space 10 to the power of negative 34 end exponent space js space straight x space 3 space straight x space 10 to the power of 8 space ms to the power of negative 1 end exponent over denominator 45 space straight x space 10 to the power of negative 9 end exponent space straight m end fraction
space equals space 4.42 space straight x space 10 to the power of negative 18 end exponent space straight j

Hence comma space the space energy space corresponds space to space light space of space wavelenght space 45 space nm space is space 4.42 space straight x space 10 to the power of negative 18 space end exponent space straight j

Question
CBSEENCH11008438

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

Solution

A.

5.79 x 106 ms-1

By Heisenberg's uncertainty principle
increment straight p space. increment straight x space greater or equal than space fraction numerator straight h over denominator 2 straight pi end fraction
or space increment straight v. increment straight x space greater or equal than fraction numerator straight h over denominator 4 πm end fraction
increment straight p space rightwards arrow space uncertainty space in space momentum
increment straight x space rightwards arrow space uncertainty space in space position
increment straight v space rightwards arrow space uncertainty space in space velocity
straight m rightwards arrow space mass space of space particle
Given comma
increment straight x space equals space 0.1 space straight A space equals space 0.1 space straight x space 10 to the power of negative 10 end exponent space straight m
straight m equals space 9.11 space straight x space 10 to the power of negative 31 end exponent space kg
straight h space equals space planck space constant space equals space 6.626 space straight x space 10 to the power of negative 34 end exponent space Js
straight pi space equals space 3.14
In space uncertain space position space increment straight v. increment straight x space equals space fraction numerator straight h over denominator 4 πm end fraction
increment straight v space straight x space 0.1 space straight x space 10 to the power of negative 10 end exponent space equals space fraction numerator 6.626 space straight x space 10 to the power of negative 34 end exponent over denominator 4 space straight x space 3.14 space straight x space 9.11 space straight x space 10 to the power of negative 31 end exponent end fraction
increment straight v space equals space fraction numerator 6.626 space straight x space 10 to the power of negative 34 end exponent over denominator 4 space straight x space 3.14 space straight x space 911 space straight x space 10 to the power of negative 31 end exponent straight x space 0.1 space straight x space 10 to the power of negative 1 end exponent end fraction ms to the power of negative 1 end exponent
equals space 5.785 space straight x space 10 to the power of 6 space ms to the power of negative 1 end exponent
5.79 space straight x space 10 to the power of 6 space ms to the power of negative 1 end exponent

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