Question

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

• A.
• B.
• C.
• D.

Answer 1

The correct option is 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, $r=\frac{a_0\times n^2}{z}$

So, for $2^{nd}$ Bohr orbit
$r$ = $\frac{a_0\times 2^2}{1}=4a_0$

KE = $\frac{1}{2}m\left(\frac{2^2{h}^2}{4{\pi }^2{m}^2\times (4a_0{)}^2}\right)$

KE = $\frac{h^2}{32{\pi }^{2}m{a}_0^2}$