Question

The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [$a_0$ is Bohr radius]

• A. $\frac{h^2}{4 \pi^2 ma_0^2}$
• B. $\frac{h^2}{16 \pi^2 ma_0^2}$
• C. $\frac{h^2}{32 \pi^2 ma_0^2}$
• D. $\frac{h^2}{64 \pi^2 ma_0^2}$

Answer 1

Answer: (C)

According to Bohr's model, $mvr=\frac{nh}{2 \pi} $
$ \Rightarrow (mv)^2=\frac{n^2 h^2}{4 \pi^2 r^2}$
$KE=\frac{1}{2} mv^2 =\frac{n^2 h^2}{8\pi^2 r^2 m} ...(i)$
Also, Bohr's radius for H-atom is, r =$n^2 \, a_0$
Substituting 'r' in Eq. (i) gives
$KE =\frac{h^2}{8 \pi^2 \, n^2 \, a_0^2 m} $
when $ n=2, KE=\frac{h^2}{32 \, \pi^2 \, a_0^2 \, m}$