Physics Part Ii Chapter 12 Atoms

Mass of electron is low as compared to proton. Hence when both enter into the uniform magnetic region, the electron will move in a circular path with higher frequency on the opposite direction to the current.

Daughter nuclei are more stable than parent nuclei. 

Bohr's postulate:

An atom has a number of stable orbits in which an electron can reside without the emission of radiant energy. Each orbit corresponds, to a certain energy level.

Electrons revolve in a circular orbit. Centripetal force is provided by electrostatic force between electron and proton.

Given as,

$\frac{{\mathrm{mv}}^2}{\mathrm{r}} = \frac{1}{4{\mathrm{\pi r}}_0}\frac{{\mathrm{e}}^2}{{\mathrm{r}}^2}$

The motion of an electron in a circular orbit is restricted in such a manner that its angular momentum is an integral multiple of h/2π, Thus

$\mathrm{L} = \mathrm{mvr} = \frac{\mathrm{nh}}{2\mathrm{\pi }}$

3.

An electron may jump spontaneously from one orbit (energy level E₁) to the other orbit (energy level E₂) (E₂ > E₁); then the energy change AE in the electron jump is given by Planck’s equation

∆E = E₂-E₁ = hv

Where h = Planck’s constant.

And v = frequency of light emitted.

When the electron jumps from nth higher orbit to pth lower orbit it emits energy in form of the photon.

E_n - E_p = hv

According to de'Broglie hypothesis

$$\begin{gathered} \mathrm{\lambda } = \frac{\mathrm{h}}{\mathrm{p}} = \frac{\mathrm{h}}{\mathrm{mv}} \\ \mathrm{n\lambda } = 2\mathrm{\pi r} \\ \mathrm{n}\frac{\mathrm{h}}{\mathrm{mv}} = 2\mathrm{\pi r} \\ \mathrm{nh} = 2\mathrm{\pi mvr} \\ \mathrm{L} = \frac{\mathrm{nh}}{2\mathrm{\pi }} \end{gathered}$$

$$\begin{gathered} {\mathrm{E}}_4 - {\mathrm{E}}_1 = \frac{\mathrm{hv}}{\mathrm{e}} \\ \frac{13.6}{4} + 13.6 = \frac{6.63 \mathrm{x} {10}^{-34} \mathrm{x} \mathrm{v}}{1.6 \mathrm{x} {10}^{-19}} \\ 13.6 = \left[1-\frac{1}{4}\right] = \frac{6.63 \mathrm{x} {10}^{-34}}{1.6 \mathrm{x} {10}^{-19}} \mathrm{v} \\ \frac{1.6 \mathrm{x} {10}^{-19} \mathrm{x} 13.6 \mathrm{x} 3}{6.63 \mathrm{x} {10}^{-34} \mathrm{x} 4} = \mathrm{v} \\ \mathrm{v} = 2.4615 \mathrm{x} {10}^{15} \mathrm{Hz} \end{gathered}$$

B.

T_n independent of n, r_n ∝ n

C.

n = 2 to n =1

C.

A.

lll

B.

${\Lambda }_n \approx A + \frac{B}{{\lambda }_n^2}$

The wavelength of emitted photon from n^th state to the ground state,

$$\begin{gathered} \frac{1}{{\Lambda }_n} = R{Z}^2 \left(\frac{1}{1^2}-\frac{1}{n^2}\right) \\ {\Lambda }_n = \frac{1}{R{Z}^2}{\left(1-\frac{1}{n^2}\right)}^{-1} \end{gathered}$$

Since n is very large, using the binomial theorem

$$\begin{gathered} {\Lambda }_n = \frac{1}{R{Z}^2}\left(1+\frac{1}{n^2}\right) \\ {\Lambda }_n = \frac{1}{R{Z}^2} + \frac{1}{R{Z}^2}\left(\frac{1}{n^2}\right) \\ As we know, {\lambda }_n = \frac{2\pi r}{n} =2\pi \left(\frac{n^2{h}^2}{4{\pi }^{2}mZ{e}^2}\right)\frac{1}{n}\propto n \\ {\Lambda }_n \approx A + \frac{B}{{\lambda }_n^2} \end{gathered}$$

A.

v_L/25

Series limit frequency of the Lyman series is given by
$$\begin{gathered} v_L = Rc{Z}^2\left(\frac{1}{1^2}-\frac{1}{{\infty }^2}\right) \\ {\nu }_L = Rc{Z}^2 ... \left(1\right) \end{gathered}$$

Series limit frequency of the Pfund series

$$\begin{gathered} v_p = Rc{Z}^2 \left(\frac{1}{5^2}-\frac{1}{{\infty }^2}\right) \\ {v}_p = \frac{v_L}{25} \end{gathered}$$

C.

4

For the last Balmer Series

B.

1:-1

In a Bohr orbit of the hydrogen atom Kinetic energy,

$$\begin{gathered} k = \frac{kz{e}^2}{2r_n} \\ Total energy, E = \frac{-kz{e}^2}{2r_n} \end{gathered}$$

so, Kinetic energy: total energy  = 1:-1

A.

$\frac{{\lambda }_0}{\left(1+{\displaystyle \frac{e{E}_0}{m{V}_0}}t\right)}$

Initial de-Brogile wavelength

${\lambda }_0 = \frac{h}{m{V}_0} .... \left(i\right)$

Acceleration of electron

$$\begin{gathered} a = \frac{e{E}_0}{m} (\because F- ma = e{E}_0) \\ Velocity after time \text{'}t\text{'} \\ V = \left(V_0+ \frac{e{E}_0}{m}t\right) \\ So , \lambda = \frac{h}{mV} = \frac{h}{m\left(V_0 + {\displaystyle \frac{e{E}_0}{m}}t\right)} \\ So, \lambda = \frac{h}{m{V}_0\left[1+{\displaystyle \frac{e{E}_0}{m{V}_0}}t\right]} = \frac{{\lambda }_0}{\left[1 + {\displaystyle \frac{e{E}_0}{m{V}_0}}t\right]}... \left(ii\right) \\ Dividing eqs \left(ii\right) by \left(i\right) \\ de - Broglie wavelength \lambda = \frac{{\lambda }_0}{1+ {\displaystyle \frac{e{E}_0}{m{V}_0}} t} \end{gathered}$$

A.

${\mathrm{r}}_{\mathrm{n}} \propto \mathrm{n}$

Given that, V = V_olog_e (r/r_o)

Field,

$$\begin{gathered} \mathrm{E} = - \frac{\mathrm{dV}}{\mathrm{dt}} = -{\mathrm{V}}_{\mathrm{o}}\left(\frac{\mathrm{r}}{{\mathrm{r}}_{\mathrm{o}}}\right) \\ \mathrm{eE} = \frac{{\mathrm{mV}}^2}{\mathrm{r}} \mathrm{or} \frac{{\mathrm{eV}}_0{\mathrm{r}}_0}{\mathrm{r}} = \frac{{\mathrm{mV}}^2}{\mathrm{r}} \\ \mathrm{V} = {\left(\frac{{\mathrm{eV}}_0{\mathrm{r}}_0}{\mathrm{m}}\right)}^{1/2} \\ \mathrm{mV} = ({\mathrm{m}}_{\mathrm{e}}{\mathrm{V}}_0{\mathrm{r}}_0{)}^{1/2} = \mathrm{constant} \\ \therefore \mathrm{mvr} = \frac{\mathrm{nh}}{2\mathrm{\pi }}, \mathrm{Bohr}\text{'}\mathrm{s} \mathrm{quantum} \mathrm{condition} \\ \mathrm{or} \mathrm{r}\propto \mathrm{n} \end{gathered}$$

C.

4 r

Radius of Bohr orbit 

${\mathrm{v}}_{\mathrm{n}} = \left(\frac{{\mathrm{n}}^2}{\mathrm{m}}\right)\frac{{\mathrm{h}}^2}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\frac{1}{\left({\displaystyle \raisebox{1ex}{$\mathrm{h}$}\!\left/ \!\raisebox{-1ex}{$2\mathrm{\pi }$}\right.}\right)}$

r ∝ n²

$$\begin{gathered} \Rightarrow \frac{{\mathrm{r}}_1}{{\mathrm{r}}_2} = {\left(\frac{{\mathrm{n}}_1}{{\mathrm{n}}_2}\right)}^2 \\ \Rightarrow \frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}={\left(\frac{1}{2}\right)}^2 \\ \Rightarrow \frac{\mathrm{r}}{{\mathrm{r}}_2} = \frac{1}{4} \\ \Rightarrow {\mathrm{r}}_2 = 4\mathrm{r} \end{gathered}$$

D.

3.59 A^o

Denoting a crystal structure in which there is an atom at each vertex and at the centre of each face of the unit cell.

lnteratomic spacing for a FCC lattice

$$\begin{gathered} \mathrm{r} = {\left[{\left(\frac{\mathrm{a}}{2}\right)}^2 + {\left(\frac{\mathrm{a}}{2}\right)}^2 + {\left(0\right)}^2\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ = \raisebox{1ex}{$\mathrm{a}$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right. \\ \mathrm{a} \mathrm{being} \mathrm{lattice} \mathrm{constant} \\ \mathrm{a} = \sqrt{2 \mathrm{r}} \\ =\sqrt{2} \times 2.54 \\ \mathrm{a} = 3.59 {\mathrm{A}}^{\mathrm{o}} \end{gathered}$$

Note:- interatomic spacing is just the nearest neighbours distance.

C.

the e/m of electrons is much greater than the e/m of protons.

According to J.J. Thomson's cathode ray tube experiment, the e/m of electrons is much greater than the e/m of protons.

D.

1215 A^o

The Rydberg formula is

$\frac{\mathrm{hc}}{{\mathrm{\lambda }}_{\mathrm{if}}} = \frac{\mathrm{m} {\mathrm{e}}^4}{8 {\mathrm{\epsilon }}_0^2 {\mathrm{h}}^2} \left(\frac{1}{{\mathrm{n}}_{\mathrm{f}}^2} - \frac{1}{{\mathrm{n}}_{\mathrm{i}}^2}\right)$

where R = $\frac{{\mathrm{me}}^4}{8{\mathrm{\epsilon }}_{\mathrm{o}}^2 {\mathrm{h}}^3\mathrm{c}}$

Wavelength of light

     $\frac{1}{\mathrm{\lambda }} = \mathrm{R} \left(\frac{1}{1^2}- \frac{1}{2^2}\right)$

Where R is Rydberg constant 

⇒    $\frac{1}{\mathrm{\lambda }} = 1.097 \times {10}^7 \times \frac{3}{4}$

∴     λ  = 1.215 × 10^-7 m

        λ  = 1215 A^o

D.

7: 108

For first line of Lyman series, 

        ${\mathrm{n}}_1 = 1 \mathrm{and} {\mathrm{n}}_2 = 2$

∴    $\frac{1}{{\mathrm{\lambda }}_1} = \mathrm{R} \left(\frac{1}{1^2} - \frac{1}{2^2}\right)$

            = R $\left(1- \frac{1}{4}\right)$

     $\frac{1}{{\mathrm{\lambda }}_1} = \frac{3\mathrm{R}}{4}$

For first line of Paschen series,

   ${\mathrm{n}}_1= 3 \mathrm{and} {\mathrm{n}}_2 = 4$

    $\frac{1}{{\mathrm{\lambda }}_2} = \mathrm{R} \left(\frac{1}{3^2} - \frac{1}{4^2}\right)$

          = R $\left(\frac{1}{9} - \frac{1}{16}\right)$

⇒         $= \frac{7\mathrm{R}}{144}$

∴    $\frac{{\mathrm{\lambda }}_1}{{\mathrm{\lambda }}_2} = \frac{7\mathrm{R}}{144} \times \frac{4}{3\mathrm{R}}$

       $\frac{{\mathrm{\lambda }}_1}{{\mathrm{\lambda }}_2} = \frac{7}{108}$

B.

10.2 V

The energy of an electron in Bohr's orbit of hydrogen atom is given by the expression

                ${\mathrm{E}}_{\mathrm{n}} = - \frac{2 {\mathrm{\pi }}^2 {\mathrm{me}}^2 {\mathrm{Z}}^2}{{\mathrm{n}}^2 {\mathrm{h}}^2 {\left( 4{\mathrm{\pi \epsilon }}_{\mathrm{o}} \right)}^2}$

                ${\mathrm{E}}_{\mathrm{n}} = -13.6 \frac{{\mathrm{Z}}^2}{{\mathrm{n}}^2} \mathrm{eV}$

Since Z = 1 for hydrogen above equation can be further simplified to

               E_n = $\frac{-13.6}{{\mathrm{n}}^2} \mathrm{eV}$

From the relation

       $∆\mathrm{E} = 13.6 \mathrm{eV} - \frac{13.6 \mathrm{eV}}{{\mathrm{n}}^2}$

             = $13.6 - \frac{13.6 \mathrm{eV}}{{\left(2\right)}^2}$

        ΔE = 10.2 eV

Therefore, excitation potential

          = $\frac{10.2}{\mathrm{e}} \mathrm{eV}$

            = 10.2 V

C.

three

            E_n  = $-\frac{13.6}{{\mathrm{n}}^2}$eV

According to Bohr's model

   E₁ = -13.6 , E₃ = -1.5

From the relation

       E₃ - E₁ = (-1.5) - (-13.6)

                   = 12.09

      E₃ - E₁ = 12.1 eV

As the energy is equal to the energy taken by n=1, n = 2,  n=3 Hence there will be three spectral line.

D.

r ∝ n²

According to Bohr's model, relation between the principal quantum number (n) and radius of
stable orbit (r) are related as 

     r_n = $\left(\frac{{\mathrm{n}}^2}{\mathrm{m}}\right) {\left(\frac{\mathrm{h}}{2\mathrm{\pi }}\right)}^2 \frac{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}{{\mathrm{e}}^2}$

⇒         r ∝ n²

C.

${\mathrm{\lambda }}_0 = \frac{2 \mathrm{mc} {\mathrm{\lambda }}^2}{\mathrm{h}}$

The de-Broglie wavelength is given by

       $\mathrm{\lambda } = \frac{\mathrm{h}}{\mathrm{p}} = \frac{\mathrm{h}}{\sqrt{2\mathrm{m} \mathrm{E}}}$                  .... (i)

where, E is the energy of the electron. The cut-off wavelength λ₀ is given by

       ${\mathrm{\lambda }}_0 = \frac{\mathrm{h} \mathrm{c}}{\mathrm{E}}$                              ....(ii)

From Eq. (i),

       λ² = $\frac{{\mathrm{h}}^2}{2 \mathrm{mE}}$

⇒     E = $\frac{{\mathrm{h}}^2}{2 \mathrm{m} {\mathrm{\lambda }}^2}$                           ....(iii)

Substituting the value of E from Eq. (ii), we get

      ${\mathrm{\lambda }}_0 = \frac{\mathrm{h} \mathrm{c}}{{\displaystyle \frac{{\mathrm{h}}^2}{2 {\mathrm{m\lambda }}^2}}}$

     λ₀ = $\frac{2 \mathrm{m} {\mathrm{c\lambda }}^2}{\mathrm{h}}$

B.

3 × 10^-10 J

Mass of proton = mass of anti-proton

                       = 1.67 × 10^-27 kg = 1amu

Energy equivalent to 1 amu = 931 MeV

So,

energy equivalent to 2 amu = 2 x 931 MeV

                                        = 1862 × 10⁶ × 1.6 × 10^-19

                                         = 2.97 × 10^-10 J

energy equivalent to 2 amu = 3 × 10^-10 J

A.

0.43

When a plane is inclined to the horizontal at an angle θ, which is greater than the angle of repose, then the body placed on the inclined plane slides down with an acceleration a.

From figure,

        R = mg cosθ                            ......(i)

Net force on the body down the inclined plane

       f = mg sinθ - F

      f = ma = mgsinθ - μR                          $\left[\because \mathrm{\mu } = \frac{\mathrm{F}}{\mathrm{R}}\right]$

∴     ma = mg sinθ - μ mg cosθ             ....[ from (i) ]

              = mg ( sinθ - μ coθ )

⇒    a = g ( sinθ - μ cosθ )

Time taken by the body to slide down the plane

       t₁ =  $\sqrt{\frac{2 \mathrm{s}}{\mathrm{a}}}$

            = $\sqrt{\frac{2\mathrm{s}}{\left( \mathrm{g} \mathrm{sin\theta } - \mathrm{\mu g} \mathrm{cos\theta } \right)}}$

       t₂ = $\sqrt{\frac{2 \mathrm{s}}{\mathrm{g} \mathrm{sin\theta }}}$                     [ in absence of friction ]

       t₁ = 2 t₂

         ${\mathrm{t}}_1^2 = 4 {\mathrm{t}}_2^2$

           ${\left| \sqrt{\frac{2 \mathrm{s}}{\left(\mathrm{g} \mathrm{sin\theta } - \mathrm{\mu g} \mathrm{cos\theta }\right)}} \right|}^2 = 4 \times {\left| \sqrt{\frac{2 \mathrm{s}}{\mathrm{g} \mathrm{sin\theta }}} \right|}^2$

⇒       $\frac{2 \mathrm{s}}{\mathrm{g} \left(\sin \mathrm{\theta } - \mathrm{\mu } \cos \mathrm{\theta }\right)} = 4 \times \frac{2\mathrm{s}}{\mathrm{g} \mathrm{sin\theta }}$

                     sinθ = 4 sinθ - 4 μ cosθ

⇒     μ =  $\frac{3}{4}$ tanθ

           = $\frac{3}{4}$ tan30^o

         μ = 0.43

D.

2.5 nm

Given:-

n₁ =2,

n₂ = 3 and 

ΔE  = 68 eV

The difference in energies of two orbits

      ΔE = 13.6 Z² $\left[\frac{1}{{\mathrm{n}}_1^2} - \frac{1}{{\mathrm{n}}_2^2}\right]$

     ΔE = 13.6 Z² $\left[\frac{1}{2^2} - \frac{1}{3^2}\right]$

    ΔE = 13.6 Z² $\left[\frac{1}{4} - \frac{1}{9}\right]$

    ΔE = 12.6 Z² $\left[\frac{9 - 4}{36}\right]$

      ΔE = 13.6 Z² $\times \frac{5}{36}$

     Z² = $\frac{68 \times 36}{13.6 \times 5}$

     Z = 6

$\because$ Wavelength of photon

     $\frac{1}{\mathrm{\lambda }}$ = Z²R $\left[\frac{1}{{\mathrm{n}}_1^2 } -\frac{1}{{\mathrm{n}}_2^2}\right]$

n₁ = 1 and n₂ = ∞

      $\frac{1}{\mathrm{\lambda }}$ = 6² × R $\left[\frac{1}{1^2} - \frac{1}{\infty }\right]$

          = 36R

    λ = $\frac{1}{36 \mathrm{R}}$

       = $\frac{1}{36 \times 1.097 \times {10}^7}$

      = $\frac{{10}^{-7}}{39.5} \mathrm{m}$

   λ = 0.025 × 10^-7 m

   λ = 2.5 nm

C.

If assertion is true but reason is false.

Assertion is true but reason is false. 

According to postulates of Bohr's atom model, the electron revolves round the nucleus in fixed orbit of definite radii. As long as the electron is in a certain orbit it does not radiate any energy. Not only the centripetal force has to be the centrifugal force, even the stable orbits are fixed by Bohr's theory.

A.

ionisation of the atom

When energy of electron in a hydrogen atom goes above 0 eV then it leaves the atom. Hence hydrogen atom becomes ionised.

D.

three

Energy in the excited state = $-$13.6 + 12.1

                                         = $-$1.5 eV

∴      $\frac{-13· 6}{{\mathrm{n}}^2}$ = $-$1.5

            n = $\sqrt{\frac{13·6}{1·5}}$

            n = 3

∴  Number of spectral lines

               = $\frac{\mathrm{n} \left(\mathrm{n} + 1\right)}{2}$

                = $\frac{3 \left(3 -1\right)}{2}$

                 = 3

C.

0.39 ma

The current gain from the Collector terminal to the Emitter terminal, I_C / I_E is called ($\mathrm{\alpha }$ )   

         $\mathrm{\alpha } = \frac{{\mathrm{I}}_{\mathrm{C}}}{{\mathrm{I}}_{\mathrm{E}}}$

  ∴         I_C = $\mathrm{\alpha }$ I_E

                  = 0$·$96 × 7$·$2

             I_C = 6$·$91 mA

Now   

           I_E = I_C + I_B

    ∴      I_B = I_E $-$ I_B

                 = $7·2 - 6·91$

            I_B = 0.29 mA