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

Answer: 3155

$mVr=\frac{nh}{2\pi }$
K.E. $=\frac{n^2{h}^2}{8{\pi }^{2}m{r}^2}$
$=\frac{4h^2}{8{\pi }^{2}m{\left(4a_0\right)}^2}$
$=\left(\frac{4}{8{\pi }^2\times 16}\right)\frac{h^2}{m{a}_0^2}$
$\Rightarrow x=315.507$
$\Rightarrow 10x=3155$ (nearest integer)