Question

Suppose the potential energy for a system of an electron and a proton separated by a distance $r$ is given by $\frac{-k{e}^2}{3r^3}$. Assume Bohr's Atomic Model is applicable. Then -

• A. Total energy in the $n^{\text{th}}$ orbit is proportional to $n^6$
• B. Total energy in the $n^{\text{th}}$ orbit is proportional to $m^{-3}$ $(m$ is mass of electron$)$
• C. Total energy in the $n^{\text{th}}$ orbit is proportional to $n^{-2}$
• D. Total energy in the $n^{\text{th}}$ orbit is proportional to $m^3$ $(m$ is mass of electron$)$

Answer 1

Answer: (A), (B)

Total energy in the $n^{\text{th}}$ orbit is proportional to $m^{-3}$ $(m$ is mass of electron$)$

The electrostatic force of attraction on the electron is responsible for providing the necessary centripetal acceleration and is given by,

$F=\frac{-dU}{dr}$

$\Rightarrow F=\frac{-d}{dr}\left[\frac{-k{e}^2}{3r^3}\right]$

$\Rightarrow F=\frac{k{e}^2}{3}\frac{d}{dr}\left(r^{-3}\right)$

$\therefore F=(-3)\frac{k{e}^2}{3}\times \left(r^{-4}\right)=\frac{-k{e}^2}{r^4}$

The negative sign shows that, force is attractive in nature.

$|F|=\frac{k{e}^2}{r^4}$

For the circular dynamics of revolving electron,

$F=\frac{m{v}^2}{r}$

$\Rightarrow \frac{k{e}^2}{r^4}=\frac{m{v}^2}{r}...\left(i\right)$

From Bohr's theory, the angular momentum,

$mvr=\frac{nh}{2\pi }...\left(ii\right)$

From Eq. $\left(i\right)$ and $\left(ii\right)$, we get,

$r=\frac{k{e}^{2}4{\pi }^2}{h^2}.\frac{m}{n^2}$

$\Rightarrow r=k_1\left(\frac{m}{n^2}\right)$

Here, $k_1$ is a constant.

Now,
Total energy $=\frac{1}{2}\times$ Potential energy

$\Rightarrow TE=\frac{1}{2}\times (-\frac{k{e}^2}{3r^3})$

$\Rightarrow TE=\frac{-k{e}^2}{6r^3}$

$\Rightarrow TE=\frac{-k{e}^2}{6{\left(\frac{k_{1}m}{n^2}\right)}^3}=\frac{-k{e}^2{n}^6}{6k_1^3{m}^3}=k_2\frac{n^6}{m^3}$

Here, $k_2$ is a constant.

$\Rightarrow TE\propto \frac{n^6}{m^3}$

$\therefore TE\propto n^6$ and $TE\propto m^{-3}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option $\left(A\right)$ and $\left(B\right)$ are correct.