Atoms and Nuclei

Ionisation potential of hydrogen $=13.6 \mathrm{~V}$

Let energy $E$ is required to excite electron to 1st excited state $n=2$

Since, $E=\frac{13.6}{n^2} Z^2 \mathrm{~eV}$

$\Rightarrow \quad E=\frac{13.6}{2^2}=\frac{13.6}{4}=3.4 \mathrm{~eV}$

As we know that in case of fussion reaction energy released is proportional to mass.

$\therefore$ When separated nucleons combine some mass gets converted to energy.

Hence, mass of stable nuclei will always be less than separated nucleons.

Given, potential energy between proton and electron,

$U=\frac{K e^2}{3 R^3}$

As we know that,

Force, $F=\frac{-d U}{d R}$

$\therefore \quad F=-\frac{K e^2}{3} \frac{d}{d R} R^{-3} \Rightarrow \frac{m v^2}{R}=-\frac{K e^2}{3} \cdot(-3) R^{-4}$

where, $m$ is mass,

$v$ is speed and $R$ is radius.

$\therefore \quad \frac{m v^2}{R}=\frac{K e^2}{R^4} $

$\Rightarrow \quad R^3=\frac{K e^2}{m v^2}$ ...... (i)

From uncertainty principle,

$\begin{aligned} L & =m v R=\frac{n h}{2 \pi} \\ \Rightarrow \quad v & =\frac{n h}{2 \pi m R} \end{aligned}$

Substituting the value in Eq. (i), we get

$\begin{aligned} & R^3=\frac{K e^2}{m \cdot \frac{n^2 h^2}{4 \pi^2 m^2 R^2}} \\ \therefore \quad & R^3=\frac{K e^2 \cdot 4 \pi^2 m^2 R^2}{m n^2 h^2} \\ \therefore \quad & R=\frac{4 \pi^2 \cdot K e^2 \cdot m}{n^2 h^2} \end{aligned}$