Question

The electric potential between a proton and an electron is given by $V=V_0\ln \frac{r}{r_0},$ where $r_0$ is a constant. Assuming Bohr's model to be applicable, Find the relation between $r_n$ and $n.$ where, $n$ being the principal quantum number.

• A. $r_n\propto n$
• B. $r_n\propto \frac{1}{n}$
• C. $r_n\propto n^2$
• D. $r_n\propto \frac{1}{n^2}$

Answer 1

The correct option is A $r_n\propto n$

Given, $V=V_0\ln \frac{r}{r_0}$

$\because E=-\frac{dV}{dr}$

$=-\frac{d}{dr}\left[V_0\ln \frac{r}{r_0}\right]=-V_0\left(\frac{r_0}{r}\right)\times \frac{1}{r_0}$

$\therefore E=-\frac{V_0}{r}$

$\Rightarrow \left|\begin{array}{c}E\end{array}\right|=\frac{V_0}{r}$

Since, bohr's atomic model is applicable, thus necessary centripetal force required for the electron is,

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

$\frac{V_0}{r}e=\frac{m{v}^2}{r}$

$\Rightarrow v=\sqrt{\frac{e{V}_0}{m}}......\left(1\right)$

From de-Broglie postulates,

$mv{r}_n=\frac{nh}{2\pi }$

$\Rightarrow v=\frac{nh}{2\pi m{r}_n}......\left(2\right)$

From (1) and (2) we get,

$\therefore r_n=\frac{nh}{2\pi \sqrt{me{V}_0}}$

$\Rightarrow r_n\propto n$

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