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

The correct option is C $\frac{h^2}{32{\pi }^{2}m{a}_0^2}$

From Bohr's theory,
$mvr=\frac{nh}{2\pi }$
Given ,
$r=4a_0$ and $n=2$

$mv\left(4a_0\right)=\frac{h}{\pi }$
so, $v=\frac{h}{4m\pi a_0}$
so $KE=\frac{1}{2}m{v}^2=\frac{1}{2}m.\frac{h^2}{16m^2{\pi }^2{a}_0^2}=\frac{h^2}{32m{\pi }^2{a}_0^2}$