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

$\frac{h^{2}}{4\pi^{2}m a_{0}^{2}}$

$\frac{h^{2}}{16 \pi^{2}ma_{0}^{2}}$

$\frac{h^{2}}{32 \pi^{2}m a_{0}^{2}}$

$\frac{h^{2}}{64 \pi^{2}ma_{0}^{2}}$

Option C

According to Bohr's model

$mvr=\frac{nh}{2\pi|}$

$\Rightarrow$ $(mv)^{2}=\frac{n^{2}h^{2}}{4\pi^{2}r^{2}}$

$\Rightarrow$ $KE=\frac{1}{2}mv^{2}=\frac{n^{2}h^{2}}{8\pi^{2}r^{2}m}$ ......(i)

Also Bohr's radius for H-atom is $r=n^{2}a_{0}$

Substiuting 'r' in equation (i) gives

$KE= \frac{h^{2}}{8 \pi^{2}n^{2}a_{0}^{2}m}$

wher n=2,

KE= $\frac{h^{2}}{32 \pi^{2}m a_{0}^{2}}$

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