Question

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

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

Answer 1

The correct option is C $\frac{h^2}{32{\pi }^{2}m{a_o}^2}$

As per Bohr's postulate,
$mvr=\frac{nh}{2\pi }$
So,
$v=\frac{nh}{2\pi mr}$
$KE=\frac{1}{2}m{v}^2$
So,
$KE=\frac{1}{2}m[\frac{nh}{2\pi mr}{]}^2$
Since $r=\frac{a_o\times n^2}{Z}$
For 2nd Bohr orbit,
$r=\frac{a_o\times 2^2}{1}=4a_o$
$KE=\frac{1}{2}m\left[\frac{2^2{h}^2}{4{\pi }^2{m}^2\times (4a_o{)}^2}\right]=\frac{h^2}{32{\pi }^{2}m{a_o}^2}$