The electrostatic force of attraction between positively charged nucleus and negatively charged electrons (Coulombian force) provides necessary centripetal force of revolution.

$$\frac{m v^2}{r_B}=-\frac{e^2}{r_B^2} \cdot \frac{1}{4 \pi \varepsilon_0}$$

By Bohr's postulates in ground state, we have

$$m v r=h$$

On solving,

$$ \begin{array}{ll} \therefore & m \frac{h^2}{m^2 r_B^2} \cdot \frac{1}{r_B}=+\left(\frac{e^2}{4 \pi \varepsilon_0}\right) \frac{1}{r_B^2} \\ \therefore & \frac{h^2}{m} \cdot \frac{4 \pi \varepsilon_0}{e^2}=r_B=0.51 \mathop A\limits^o\quad\text{[This is Bohr's radius]} \end{array}$$

$$\begin{aligned} &\text { The potential energy is given by }\\ &\begin{aligned} -\left(\frac{e^2}{4 \pi r_0}\right) \cdot \frac{1}{r_B} & =-27.2 \mathrm{eV} ; \mathrm{KE}=\frac{m v^2}{2} \\ & =\frac{1}{2} m \cdot \frac{h^2}{m^2 r_B^2}=\frac{h}{2 m r_B^2}=+13.6 \mathrm{eV} \end{aligned} \end{aligned}$$

Now, for an spherical nucleus of radius $R$,

If $R

If $R>>r_B$ the electron moves inside the sphere with radius $r_B^{\prime}\left(r_B^{\prime}=\right.$ new Bohr radius).

$$\begin{aligned} &\begin{aligned} \text{Charge inside}\quad r_B^{\prime 4} & =e\left(\frac{r_B^{\prime 3}}{R^3}\right) \\ \therefore\quad r_B^{\prime} & =\frac{h^2}{m}\left(\frac{4 \pi \varepsilon_0}{e^2}\right) \frac{R^3}{r_B^{\prime 3}} \\ r_B^{\prime 4} & =(0.51 \mathop A\limits^o) \cdot R^3 \quad [R=10 \mathop A\limits^o]\\ & =510(\mathop A\limits^o)^4 \end{aligned}\\ \end{aligned}$$

$$\begin{aligned} \therefore r_B^{\prime} \simeq(510)^{1 / 4} \mathop A\limits^o & < R \\ \mathrm{KE} & =\frac{1}{2} m v^2=\frac{m}{2} \cdot \frac{h}{m^2 r_B^{\prime 2}}=\frac{h}{2 m} \cdot \frac{1}{r_B^{\prime 2}} \\ & =\left(\frac{h^2}{2 m r_B^2} \cdot\left(\frac{r_B^2}{r_B^{\prime 2}}\right)=(13.6 \mathrm{eV}) \frac{(0.51)^2}{(510)^{1 / 2}}=\frac{3.54}{22.6}=0.16 \mathrm{eV}\right. \\ \mathrm{PE} & =+\left(\frac{e^2}{4 \pi \varepsilon_0}\right) \cdot\left(\frac{r_B^{\prime 2}-3 R^2}{2 R^3}\right) \\ & =+\left(\frac{e^2}{4 \pi \varepsilon_0} \cdot \frac{1}{r_B}\right) \cdot\left(\frac{r_B\left(r_B^{\prime 2}-3 R^2\right)}{R^3}\right)=+(27.2 \mathrm{eV})\left[\frac{0.51(\sqrt{510}-300)}{1000}\right] \\ & =+(27.2 \mathrm{eV}) \cdot \frac{-141}{1000}=-3.83 \mathrm{~eV} \end{aligned}$$

The energy of the $n$th state $E_n=-Z^2 R \frac{1}{n^2}$ where $R$ is the Rydberg constant and $Z=24$.

The energy released in a transition from 2 to 1 is $\Delta E=Z^2 R\left(1-\frac{1}{4}\right)=\frac{3}{4} Z^2 R$.

The energy required to eject a $n=4$ electron is $E_4=Z^2 R \frac{1}{16}$. Thus, the kinetic energy of the Auger electron is

$$\begin{aligned} \mathrm{KE} & =Z^2 R\left(\frac{3}{4}-\frac{1}{16}\right)=\frac{1}{16} Z^2 R \\ & =\frac{11}{16} \times 24 \times 24 \times 13.6 \mathrm{~eV} \\ & =5385.6 \mathrm{~eV} \end{aligned}$$

$$\begin{aligned} &\text { For } m_p=10^{-6} \text { times, the mass of an electron, the energy associated with it is given by }\\ &\begin{aligned} m_p c^2 & =10^{-6} \times \text { electron mass } \times c^2 \\ & \approx 10^{-6} \times 0.5 \mathrm{MeV} \\ & \approx 10^{-6} \times 0.5 \times 1.6 \times 10^{-13} \\ & \approx 0.8 \times 10^{-19} \mathrm{~J} \end{aligned} \end{aligned}$$

$$\begin{aligned} &\text { The wavelength associated with is given by }\\ &\begin{aligned} \frac{\hbar}{m_p c} & =\frac{\hbar c}{m_p c^2}=\frac{10^{-34} \times 3 \times 10^8}{0.8 \times 10^{-19}} \\ & \approx 4 \times 10^{-7} \mathrm{~m} \gg \text { Bohr radius } \\ |\mathbf{F}| & =\frac{e^2}{4 \pi \varepsilon_0}\left[\frac{1}{r^2}+\frac{\lambda}{r}\right] \exp (-\lambda r) \end{aligned} \end{aligned}$$

where, $\quad \lambda^{-1}=\frac{\hbar}{m_p \mathrm{c}} \approx 4 \times 10^{-7} \mathrm{~m} \gg r_B$

$\therefore \quad \lambda \ll \frac{1}{r_B}$ i.e., $\lambda r_B \ll 1$

$$\begin{aligned} U(r) & =-\frac{e^2}{4 \pi \varepsilon_0} \cdot \frac{\exp (-\lambda r)}{r} \\ m v r & =\hbar \quad \therefore \quad v=\frac{\hbar}{m r} \\ \text{Also,}\quad\frac{m v^2}{r} & =\approx\left(\frac{e^2}{4 \pi \varepsilon_0}\right)\left[\frac{1}{r^2}+\frac{\lambda}{r}\right] \\ \therefore\quad\frac{\hbar^2}{m r^3} & =\left(\frac{e^2}{4 \pi \varepsilon_0}\right)\left[\frac{1}{r^2}+\frac{\lambda}{r}\right] \\ \therefore\quad\frac{\hbar^2}{m} & =\left(\frac{e^2}{4 \pi \varepsilon_0}\right)\left[r+\pi r^2\right. \\ \text{If } \lambda=0;\quad r & =r_B=\frac{\hbar}{m} \cdot \frac{4 \pi \varepsilon_0}{e^2} \\ \frac{\hbar^2}{m} & =\frac{e^2}{4 \pi \varepsilon_0} \cdot r_B \end{aligned}$$

Since, $\lambda^{-1} \gg r_B$, put $r=r_B+\delta$

$\therefore \quad r_B=r_B+\delta+\lambda\left(r_B^2+\delta^2+2 \delta r_B\right) ;$ neglect $\delta^2$

or $\quad 0=\lambda r_B^2+\delta\left(1+2 \lambda r_B\right)$

$\delta=\frac{-\lambda r_B^2}{1+2 \lambda r_B} \approx \lambda r_B^2\left(1-2 \lambda r_B\right)=-\lambda r_B^2$

$$\begin{aligned} &\text { Since, } \lambda r_B \ll 1\\ \therefore\quad &V(r)=-\frac{e^2}{4 \pi \varepsilon_0} \cdot \frac{\exp \left(-\lambda \delta-\lambda r_B\right)}{r_B+\delta} \end{aligned}$$

$\therefore \quad \quad V(r)=-\frac{e^2}{4 \pi \varepsilon_0} \frac{1}{r_B}\left[\left(1-\frac{\delta}{r_B}\right) \cdot\left(1-\lambda r_B\right)\right]$

$$\begin{aligned} & \cong(-27.2 \mathrm{eV}) \text { remains unchanged } \\ \mathrm{KE} & =-\frac{1}{2} m v^2=\frac{1}{2} m \cdot \frac{h^2}{m r^2}=\frac{h^2}{2\left(r_B+\delta\right)^2}=\frac{h^2}{2 r_B^2}\left(1-\frac{2 \delta}{r_B}\right) \\ & =(13.6 \mathrm{eV})\left[1+2 \lambda r_B\right] \end{aligned}$$

$$\begin{aligned} &\begin{aligned} \text { Total energy }\quad & =-\frac{e^2}{4 \pi \varepsilon_0 r_B}+\frac{h^2}{2 r_B^2}\left[1+2 \lambda r_B\right] \\ & =-27.2+13.6\left[1+2 \lambda r_B\right] \mathrm{eV} \\ \text { Change in energy }\quad& =13.6 \times 2 \lambda r_B \mathrm{eV}=27.2 \lambda r_B \mathrm{~eV} \end{aligned}\\ \end{aligned}$$

Considering the case, when $r \leq R_0=1 \mathop A\limits^o$

$$\begin{aligned} & \text { Let } \varepsilon=2+\delta \\ & F=\frac{q_1 q_2}{4 \pi \varepsilon_0} \cdot \frac{R_0^\delta}{r^{2+\delta}} \end{aligned}$$

where,

$$\begin{aligned} \frac{q_1 q_2}{4 \pi_0 \varepsilon_0} & =\left(1.6 \times 10^{-19}\right)^2 \times 9 \times 10^9 \\ & =23.04 \times 10^{-29} \mathrm{~N} \mathrm{~m}^2 \end{aligned}$$

The electrostatic force of attraction between positively charged nucleus and negatively charged electrons (Coulombian force) provides necessary centripetal force.

$$\begin{aligned} & =\frac{m v^2}{r} \quad \text { or } \quad v^2=\frac{\wedge R_0^\delta}{m r^{1+\delta}} \quad\text{.... (i)}\\ m v r & =n h \cdot r=\frac{n \hbar}{m v}=\frac{n h}{m}\left[\frac{m}{\wedge R_0^\delta}\right]^{1 / 2} r^{1 / 2+\delta / 2}\quad \text{[Applying Bohr's second postulates]} \end{aligned}$$

Solving this for $r$, we get $\quad r_n=\left[\frac{n^2 \hbar^2}{m \wedge R_0^\delta}\right]^{\frac{1}{1-\delta}}$ where, $r_n$ is radius of $n$th orbit of electron.

For $n=1$ and substituting the values of constant, we get

$$\begin{aligned} r_1 & =\left[\frac{\hbar^2}{m \wedge R_0^\delta}\right]^{\frac{1}{1-\delta}} \\ r_1 & =\left[\frac{1.05^2 \times 10^{-68}}{9.1 \times 10^{-31} \times 2.3 \times 10^{-28} \times 10^{+19}}\right]^{\frac{1}{2.9}} \\ & =8 \times 10^{-11} \\ & =0.08 \mathrm{~nm}\quad (< 0.1 \text{nm}) \end{aligned}$$

This is the radius of orbit of electron in ground state of hydrogen atom.

$$\begin{aligned} v_n & =\frac{n \hbar}{m r_n}=n \hbar\left(\frac{m \wedge R_0^\delta}{n^2 \hbar^2}\right)^{\frac{1}{1-\delta}} \\ \text { For } n & =1, v_1-\frac{\hbar}{m r_1}=1.44 \times 10^6 \mathrm{~m} / \mathrm{s} \end{aligned}$$

[This is the speed of electron in ground state]

$$\begin{aligned} &\mathrm{KE}=\frac{1}{2} m v_1^2-9.43 \times 10^{-19} \mathrm{~J}=5.9 \mathrm{eV}\\ &\text { [This is the KE of electron in ground state] } \end{aligned}$$

PE till $R_0=-\frac{\Lambda}{R_0}$ [This is the PE of electron in ground state at $r=R_0$ ]

PE from $R_0$ to $r=+\wedge R_0^\delta \int_{R_0}^r \frac{d r}{r^{2+\delta}}=+\frac{\wedge R_0^\delta}{-1-\delta}\left[\frac{1}{r^{1+\delta}}\right]_{R_0}^r$

[This is the PE of electron in ground state at $R_0$ to $r$ ]

$$\begin{aligned} & =-\frac{\wedge R_0^\delta}{1+\delta}\left[\frac{1}{r^{1+\delta}}-\frac{1}{R_0^{1+\delta}}\right]=-\frac{\Lambda}{1+\delta}\left[\frac{R_0^\delta}{r^{1+\delta}}-\frac{1}{R_0}\right] \\ P E & =-\frac{\Lambda}{1+\delta}\left[\frac{R_0^\delta}{r^{1+\delta}}-\frac{1}{R_0}+\frac{1+\delta}{R_0}\right] \\ P E & =-\frac{\Lambda}{-0.9}\left[\frac{R_0^{-1.9}}{r^{-0.9}}-\frac{1.9}{R_0}\right] \\ & =\frac{2.3}{0.9} \times 10^{-18}\left[(0.8)^{0.9}-1.9\right] \mathrm{J}=-17.3 \mathrm{̃eV} \end{aligned}$$

Total energy is $(-17.3+5.9)=-11.4 \mathrm{eV}$

This is the required TE of electron in ground state.