In a hydrogen atom,the binding energy of the | vedclass

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(A) We know that for a hydrogen atom,the kinetic energy $(K)$ is equal to the binding energy $(E_n)$: $K = \frac{1}{2} m v^2 = E_n$ --- $(i)$ Accordin

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$2E_n / nh$

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(A) We know that for a hydrogen atom,the kinetic energy $(K)$ is equal to the binding energy $(E_n)$: $K = \frac{1}{2} m v^2 = E_n$ --- $(i)$ According to Bohr's quantization condition of angular momentum: $mvr = \frac{nh}{2\pi}$ --- $(ii)$ From $(i)$,we have $v^2 = \frac{2E_n}{m}$,so $v = \sqrt{\frac{2E_n}{m}}$. From $(ii)$,the radius $r = \frac{nh}{2\pi mv}$. The frequency of revolution $(f)$ is given by $f = \frac{v}{2\pi r}$. Substituting the value of $r$ from $(ii)$ into the frequency formula: $f = \frac{v}{2\pi (nh / 2\pi mv)} = \frac{v^2 m}{nh} = \frac{(2E_n / m) m}{nh} = \frac{2E_n}{nh}$. Therefore,the frequency of revolution is $\frac{2E_n}{nh}$.

In a hydrogen atom,the binding energy of the electron in the $n^{th}$ state is $E_n$. Then,the frequency of revolution of the electron in the $n^{th}$ orbit is:

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