Q. The electric potential between a proton and an electron is given by $V = V_{0}$ In$\ \frac{r}{r_{0}}$ , where $r_{0}$ is a constant. Assuming Bohr’s model to be applicable, write variation of $r_{n}$ with $n,\ n$ being the principal quantum number?

A) $r_{n} \propto \ n$
B) $r_{n} \propto \frac{1}{n}$
C) $r_{n} \propto \ n^{2}$
D) $r_{n} \propto \frac{1}{n^{2}}$
Explanation

(a) $\ \ \ \ \ \ \ U\ = eV\ = eV_{0}\ln\left( \frac{r}{r_{0}} \right)$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |F| = \ \left| - \frac{dU}{dr} \right| = \frac{{eV}_{0}}{r}$
This force will provide the necessary centripetal force. Hence
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{{mv}^{2}}{r} = \frac{{eV}_{0}}{r}$
or $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v$= $\sqrt{\frac{{eV}_{0}}{m}}$ ….(i)
Moreover
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ mvr\ = \frac{nh}{2\pi}$ …..(ii)
Dividing Eq. (ii) by Eq. (i), we have
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ mr = \left( \frac{nh}{2\pi} \right)\sqrt{\frac{\ m}{{eV}_{0}}}$
Or $r_{n} \propto n$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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