The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [ $a_{0}$ is Bohr radius]:

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$\frac{h^{2}}{4 π^{2} m a_{0}^{2}}$

$\frac{h^{2}}{16 π^{2} m a_{0}^{2}}$

$\frac{h^{2}}{32 π^{2} m a_{0}^{2}}$

$\frac{h^{2}}{64 π^{2} m a_{0}^{2}}$

answer is C.

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Detailed Solution

As per Bohr’s postulate

$\begin{array}{l} m v r = \frac{n h}{2 x} \\ S o, v = \frac{n h}{2 x m r} \\ K E = \frac{1}{2} m (\frac{n h}{2 x m r}) \end{array}$

$\begin{array}{l} S o, K E = \frac{1}{2} m {(\frac{n h}{2 x, m r})}^{2} \\ S i n c e, r = \frac{a_{x} \times n^{2}}{z} \end{array}$

So, for 2^nd Bohr orbit

$r = \frac{a_{x} \times 2^{2}}{1} = 4 a_{x}$

$\begin{array}{l} K E = \frac{1}{2} m (\frac{2^{2} h^{2}}{4 x^{2} m^{2}, {(4 a_{x})}^{2}}) \\ K E = \frac{h^{2}}{32 π^{2} m {a_{0}}^{2}} \end{array}$

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