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

We know that,

$E\propto \frac{Z^2}{n^2}$

$r_n$ = $\frac{r_{1}forH\times n^2}{Z}$

$r_n\propto n^2$

$r_n\propto (2{)}^2{a}_0$

$r_n\propto 4a_0$

$mvr=\frac{nh}{2\pi }=\frac{2}{2}\frac{h}{\pi }$

$\therefore v=\frac{h}{\pi mr}(r=4a_0)$

= $\frac{h}{\pi m\times 4a_0}$

So, KE = $\frac{1}{2}m{v}^2$

= $\frac{1}{2}m{\left(\frac{h}{4\pi \times m{a}_0}\right)}^2=\frac{h^2}{32{\pi }^{2}m{a}_0^2}$