Question

The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is equal to $\frac{h^2}{xm{a}_0^2}$. The value of 10x is .
($a_0$ is radius of Bohr's orbit) (Nearest integer)
[Given: $\pi =3.14$]

Answer 1

Kinetic energy of an electron in nth orbit of Bohr atom :
$\frac{1}{2}m{v}^2=\frac{(mv{)}^2}{2m}$
In Bohr's model,
$mvr=\frac{nh}{2\pi }$ or
$mv=\frac{nh}{2\pi r}$
$KE=\frac{n^2{h}^2}{8{\pi }^{2}m{r}^2}$
For 2nd orbit of H-atom
$n=2\text{and}r=4a_0$
$\therefore KE=\frac{4h^2}{8{\pi }^{2}m\times 16a_0^2}=\frac{h^2}{315.5m{a}_0^2}$

$\therefore x=315.5;10x=3155$