The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [a0 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

We know that;

$m v r = \frac{n h}{2 π} \Rightarrow P = \frac{n h}{2 π r}; K E = \frac{P^{2}}{2 m} = \frac{n^{2} h^{2}}{8 π^{2} m \cdot r^{2}}$ Also, $r = a_{0} \times n^{2} \Rightarrow KE = \frac{h^{2}}{8 π^{2} m n^{2} a_{0}^{2}}$ ; Put n=2

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