Question

The kinetic energy of an electron in the second Bohr's orbit of a hydrogen atom is:
[$a_0$ is Bohr's 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 $\frac{h^2}{32{\pi }^{2}m{a}_0^2}$

$K.E.=\frac{1}{2}m{v}^2....\left(1\right)$
$mvr=\frac{nh}{2\pi }\to$ (Bohr's model)
$(mv{)}^2=\frac{n^2{h}^2}{4{\pi }^2{r}^2}$
$m{v}^2=\frac{1}{m}\times \frac{n^2{h}^2}{4{\pi }^2{r}^2}....\left(2\right)$
Put (2) in (1)
$K.E\Rightarrow \frac{1}{2}\times \frac{1}{m}\times \frac{n^2{h}^2}{4{\pi }^2{r}^2}$
Now, $n_o=2,r_1=a_o,r_2=a_o\times (2{)}^2=4a_0$.
$K.E.\Rightarrow \frac{1}{2}\times \frac{2^2{h}^2}{4{\pi }^2(4a_o{)}^{2}m}$
$\Rightarrow \frac{h^2}{32{\pi }^2{a}_o^{2}m}$