Question

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

• A. $\frac{h^2}{4{\pi }^{2}m{a}_0^2}$
• B. $\frac{h^2}{16{\pi }^{2}m{a}_0^2}$
• C. $\frac{h^2}{32{\pi }^{2}m{a}_0^2}$
• D. $\frac{h^2}{64{\pi }^{2}m{a}_0^2}$

Answer 1

Answer: (C)

According to Bohr's model, $mvr=\frac{nh}{2\pi }$
$\Rightarrow (mv{)}^2=\frac{n^2{h}^2}{4{\pi }^2{r}^2}$
$KE=\frac{1}{2}m{v}^2=\frac{n^2{h}^2}{8{\pi }^2{r}^{2}m}...(i)$
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 $n=2,KE=\frac{h^2}{32{\pi }^2{a}_0^{2}m}$