Physics Ncert Solutions Class 12th

12.14 Charge of an electron, e = 1.6 $*{10}^{-19}$ C

Mass of an electron, $m_e$ = 9.1 $*{10}^{-31}$ kg

Speed of light, c = 3 $*{10}^8$ m/s

Let us take a quantity involving the given quantities as $\left(\frac{e^2}{4?{?}_0{m}_e{c}^2}\right)$

Where

${?}_0$ = permittivity of free space and

$\frac{1}{4?{?}_0}$ = 9.1 $*{10}^9$ N $m^2{C}^{-1}$

Hence, $\left(\frac{e^2}{4?{?}_0{m}_e{c}^2}\right)$ = 9.1 $*{10}^9*\frac{{(1.6\mathrm{}*{10}^{-19})}^2}{9.1\mathrm{}*{10}^{-31}*{(3\mathrm{}*{10}^8)}^2}$ = 2.844 $*{10}^{-15}$ m

Hence, the numerical value of the taken quantity is much smaller than the typical size of an atom.

Charge of an electron, e = 1.6 $*{10}^{-19}$ C

Mass of an electron, $m_e$ = 9.1 $*{10}^{-31}$ kg

Planck's constant, h = 6.623 $*{10}^{-34}$ Js

Hence, the value of the quantity taken is in the order of the atomic size.

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.