Class 12th

12.13 It is given that a hydrogen atom de-excites from an upper level (n) to a lower level (n-1)

We have the relation for energy ( $E_1$ ) of radiation at level n as:

$E_1$ = h ${?}_1$ = $\frac{hm{e}^2}{{\left(4?\right)}^2{?}_0^2({\frac{h}{2?})}^2}$ $*\frac{1}{{\left(n\right)}^2}$ ………………(i)

Where,

${?}_1$ = Frequency of radiation at level n

h = Planck's constant

m = mass of hydrogen atom

e = charge of an electron

${?}_0$ = Permittivity of free space

Now, the relation for energy ( $E_2$ ) of radiation at level (n-1) is given as:

$E_2$ = h ${?}_2$ = $\frac{hm{e}^2}{{\left(4?\right)}^2{?}_0^2({\frac{h}{2?})}^2}$ $*\frac{1}{{(n-1)}^2}$ ………………(ii)

Where,

${?}_2$ = Frequency of radiation at level (n-1)

Energy (E) released as a result of de-excitation:

E = $E_2-E_1$

$h?$ = $E_2-E_1$ …………………………(iii)

where,

$?$ = Frequency of radiation emitted

Putting values of equation (i) and (ii) in equation (iii), we get,

$?$ = $\frac{m{e}^2}{{\left(4?\right)}^2{?}_0^2({\frac{h}{2?})}^2}$ $\left[\frac{1}{{(n-1)}^2}-\frac{1}{{\left(n\right)}^2}\right]$

= $\frac{m{e}^2(2n-1)}{{\left(4?\right)}^2{?}_0^2({\frac{h}{2?})}^2{n}^2{(n-1)}^2}$

For large n, we can write (2n-1) = 2n and (n-1) = n

Therefore, $?$ = $\frac{m{e}^2}{32{?}^3{?}_0^2({\frac{h}{2?})}^3{n}^3}$ ……….(iv)

Classical relation of frequency of revolution of an electron is given as:

${?}_c$ = $\frac{v}{2?r}$ ………………….(v)

Velocity of the electron in the $n^{th}$ orbit is given as:

v = $\frac{e^2}{4?{?}_0\left(\frac{h}{2?}\right)n}$ …………….(vi)

And, radius of the $n^{th}$ orbit is given as:

r = $\frac{4?{?}_0({\frac{h}{2?})}^2}{m{e}^2}{n}^2$ ……………….(vii)

By putting the values of (vi) and (vii) in equation (v), we get

${?}_c$ = $\frac{m{e}^2}{32{?}^3{?}_0^2({\frac{h}{2?})}^3{n}^3}$

Hence, the frequency of radiation emitted by the hydrogen atom is equal to its classical orbital frequency.

12.12 Radius of the first Bohr orbit is given by the relation:

$r_1$ = $\frac{4?{?}_0{\left(\frac{h}{2?}\right)}^2}{m_e{e}^2}$ ………………(1)

Where,

${?}_0=$ Permittivity of free space

h = Planck's constant = 6.626 $*{10}^{-34}$ Js

$m_e$ = mass of electron = 9.1 $*{10}^{-31}$ kg

e = Charge of electron = 1.9 $*{10}^{-19}$ C

$m_p$ = mass of a proton = 1.67 $*{10}^{-27}$ kg

r = distance between the electron and proton

Coulomb attraction between an electron and a proton is given as:

$F_c$ = $\frac{e^2}{4?{?}_0{r}^2}$ ……………….(2)

Gravitational force of attraction between an electron and a proton is given as:

$F_G$ = $\frac{G{m}_p{m}_e}{r^2}$ ……………….(3)

Where, G = Gravitational constant = 6.67 $*{10}^{-11}$ N $m^2$ / ${kg}^2$

If the electrostatic (Coulomb) force and the gravitational force between an electron and a proton are equal, then we can write

$F_c=F_G$

$\frac{e^2}{4?{?}_0{r}^2}=\frac{G{m}_p{m}_e}{r^2}$

or

$G{m}_p{m}_e=\frac{e^2}{4?{?}_0}$

By putting the above value in equation (1), we can write

$r_1$ = $\frac{4?{?}_0}{e^2}*\frac{{\left(\frac{h}{2?}\right)}^2}{m_e}$ = $\frac{1}{G{m}_p{m}_e}$ $*\frac{{\left(\frac{h}{2?}\right)}^2}{m_e}$ = $\frac{{\left(\frac{h}{2?}\right)}^2}{G{m}_p{m}_e^2}$

= $\frac{{\left(\frac{6.626\mathrm{}*{10}^{-34}}{2?}\right)}^2}{6.67\mathrm{}*{10}^{-11}*1.67\mathrm{}*{10}^{-27}*{9.1\mathrm{}*{10}^{-31}}^2}$ = $\frac{1.112*{10}^{-68}}{9.224*{10}^{-98}}$ = 1.21 $*{10}^{29}$ m

It is known that the universe is 156 billion light years wide or 1.5 $*{10}^{27}$ m wide. Hence, we can conclude that the radius of the first Bohr orbit is much greater than the estimated size of the whole universe.

12.11 About the same - The average angle of deflection $?$ -particles by a thin gold foil predicted by Thomson's model is about the same size as predicted by Rutherford's model. This is because the average angle was taken in both models.

Much less - The probability of scattering of $?$ -particles at angles greater than 90 $°$ predicted by Thomson's model is much less than that predicted by Rutherford's model.

Scattering is mainly due to single collisions. The chances of a single collision increases linearly with the number of target atoms. Since the number of target atoms increase with an increase in thickness, the collision probability depends linearly on the thickness of the target.

Thomson's model - It is wrong to ignore multiple scattering in Thomson's model for the calculation of average angle of scattering of $?$ -particles by a thin foil. This is because a single-collision causes very little deflection in this model. Hence, the observed average scattering angle can be explained only by considering multiple scattering.

12.9 It is given that the energy of the electron beam used to bombard gaseous hydrogen at room temperature is 12.5 eV. Also, the energy of the gaseous hydrogen in its ground state at room temperature is – 13.6 eV.

When gaseous hydrogen is bombarded with an electron beam, the energy of the gaseous hydrogen becomes -13.6 + 12.5 eV i.e. -1.1 eV

Orbital energy is related to orbit level(n) as:

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

For n = 3, E = $\frac{-13.6}{9}$ eV = - 1.5 eV

This energy is approximately equal to the energy of the gaseous hydrogen. It can be concluded that the electron has jumped from n = 1 to n = 3 level.

During the de-excitation, the electron can jump from n=3 to n=1 directly, which forms a line of the Lyman series of hydrogen spectrum.

We have the relation for wave number for Lyman series as:

$\frac{1}{?}$ = $R_y\left(\frac{1}{1^2}-\frac{1}{n^2}\right)$ where

$R_y$ = Rydberg constant = 1.097 $*{10}^7{m}^{-1}$

$?=$ Wavelength of radiation emitted by the transition of electron

For n = 3, we get

$\frac{1}{?}$ = 1.097 $*{10}^7*\left(\frac{1}{1^2}-\frac{1}{3^2}\right)$

$?=1.215*{10}^{-7}m$ = 102.6 nm

If the electron jumps from n = 2 to n = 1, then the wavelength of the radiation is given as:

$\frac{1}{?}$ = 1.097 $*{10}^7*\left(\frac{1}{1^2}-\frac{1}{2^2}\right)$

$?=1.215*{10}^{-7}m$ = 121.5 nm

If the transition takes place from n = 3 to n = 2, then the wavelength is given as:

$\frac{1}{?}$ = 1.097 $*{10}^7*\left(\frac{1}{2^2}-\frac{1}{3^2}\right)$

$?=6.563*{10}^{-7}m$ = 656.3 nm

This radiation corresponds to the Balmer series of the hydrogen spectrum.

Hence, in Lyman series, two wavelengths i.e. 102.6 nm and 121.5 nm are emitted. And in the Balmer series, one wavelength i.e. 656.3 nm is emitted.

12.7 Let $v_1$ be the speed of the electron of the hydrogen atom in the ground state level, $n_1=1.$ For charge e of an electron, $v_1$ is given by the relation

$v_1=\frac{e^2}{n_{1}4?{?}_0\left(\frac{h}{2?}\right)}$ = $\frac{e^2}{2n_1{?}_{0}h}$

$where$ , e = 1.6 $*{10}^{-19}$ C 

${?}_0=$ Permittivity of free space = 8.85 $*{10}^{-12}$ $N^{-1}{C}^2{m}^{-2}$