Question

The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is:
($a_0$ is bohr radius)​

• A. $\frac{h^2}{4{\pi }^{2}m{a}_o^2}$
• B. $\frac{h^2}{16{\pi }^{2}m{a}_o^2}$
• C. $\frac{h^2}{32{\pi }^{2}m{a}_o^2}$
• D. $\frac{h^2}{64{\pi }^{2}m{a}_o^2}$

Answer 1

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

We know,
$K.E.=\frac{1}{2}m{v}^2...\left(i\right)$
According to Bohr's model
$mvr=\frac{nh}{2\pi }$
$\Rightarrow (mv{)}^2$ = $\frac{n^2{h}^2}{4{\pi }^2{r}^2}$

$\Rightarrow m{v}^2$ = $\frac{1}{m}\times \frac{n^2{h}^2}{4{\pi }^2{r}^2}...\left(ii\right)$

Putting value of $\left(ii\right)$ in $\left(i\right)$
$K.E.=\frac{1}{2}\times \frac{1}{m}\times \frac{n^2{h}^2}{4{\pi }^2{r}^2}$

Now, $n=2$, $r_1=a_o$, $r_2=a_o\times 2^2=4a_o$

$\therefore K.E.=\frac{1}{2}\times \frac{1}{m}\times \frac{2^2{h}^2}{4{\pi }^2(4a_o{)}^2}$

$\Rightarrow K.E.=\frac{h^2}{32{\pi }^{2}m{a}_o^2}$