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.