Question

The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is equal to $\frac{ h ^{2}}{ xma _{0}^{2}}$. The value of $10 x$ is____ ( $a_{0}$ is radius of Bohr's orbit)
(Nearest integer) [Given : $\pi=3.14$ ]

Answer 1

$mVr =\frac{ nh }{2 \pi}$
K.E. $=\frac{ n ^{2} h ^{2}}{8 \pi^{2} mr ^{2}}$
$=\frac{4 h ^{2}}{8 \pi^{2} m \left(4 a _{0}\right)^{2}}$
$=\left(\frac{4}{8 \pi^{2} \times 16}\right) \frac{ h ^{2}}{ ma _{0}^{2}}$
$\Rightarrow x =315.507$
$\Rightarrow 10 x =3155$ (nearest integer)