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 nth orbit is:

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$\frac{n h}{2 π} (\frac{K}{m})$

$\frac{n h}{2 π} {(\frac{K}{m})}^{\frac{1}{2}}$

$n h (\frac{K}{m})$

$\frac{n h}{2 π} {(\frac{m}{K})}^{\frac{1}{2}}$

answer is B.

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Detailed Solution

Given
Potential energy of an electron $U = \frac{K r^{2}}{2}$
Where K=Positive constant
We use force on electron
$\begin{array}{l} F = \frac{d U}{d r} = K r = \frac{m v^{2}}{r} (∵ F_{c} = \frac{m v^{2}}{r}) \\ \Rightarrow m v^{2} = K r^{2} and m v r = \frac{n h}{2 π} \\ m r (\sqrt{\frac{K}{m}} r) = \frac{n h}{2 π} \Rightarrow r = \sqrt{\frac{n h}{2 π \sqrt{K m}}} \end{array}$

Energy of n^thorbit is

$\begin{array}{l} E_{n} = \frac{K r^{2}}{2} + \frac{K r^{2}}{2} = K r^{2} \\ \Rightarrow E_{n} = \frac{K n h}{2 π \sqrt{K m}} = \frac{n h}{2 π} \sqrt{\frac{K}{m}} \end{array}$