Question

In Bohr's theory the potential energy of an electron at a position is $\frac{K r^{2}}{2}$ (where $K$ is a positive constant); then the quantized energy of the electron in $n$th orbit is :

• A. $\frac{n h}{2 \pi}\left(\frac{K}{m}\right)$
• B. $\frac{n h}{2 \pi}\left(\frac{K}{m}\right)^{1 / 2}$
• C. $n h\left(\frac{K}{m}\right)$
• D. $\frac{n h}{2 \pi}\left(\frac{m}{K}\right)^{1 / 2}$

Answer 1

Answer: (B)

We use force on electron
$F =\frac{d U}{d r}=K r=\frac{m v^{2}}{r}$
$\Rightarrow m v^{2} =K r^{2}$
and $m v r =\frac{r h}{2 \pi}$
$m r\left(\sqrt{\frac{K}{m}} r\right) =\frac{n h}{2 \pi}$
$\Rightarrow r =\sqrt{\frac{n h}{2 \pi \sqrt{K m}}}$
Energy of $n^{\text {th }}$ orbit is
$E_{n}=\frac{K r^{2}}{2}+\frac{K r^{2}}{2}=K r^{2}$
$\Rightarrow E_{n}=\frac{K n h}{2 \pi \sqrt{K m}}=\frac{n h}{2 \pi} \sqrt{\frac{K}{m}}$