1)

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


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

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

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

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

Answer:

Option C

Explanation:

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