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)

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, $KE=\frac{1}{2}m{\left(\frac{nh}{2\pi mr}\right)}^2$
Since, $r=\frac{a_o\times n^2}{z}$
So, for $2^{\text{nd }}$ Bohr orbit
$r=\frac{a_0\times 2^2}{1}=4a_a$
$KE=\frac{1}{2}m\left(\frac{2^2{h}^2}{4{\pi }^2{m}^2\times {\left(4a_0\right)}^2}\right)$
$KE=\frac{h^2}{32{\pi }^{2}m{a}_0^2}$