Question

The velocity of $e^{-}$ in a certain Bohr orbit of the hydrogen atom bears the ratio 1:275 to the velocity of light. What is the quantum no. "n" of the orbit and the wave no. of the radiation emitted for the transition from the quantum state (n+1) to the ground state.

Answer 1

$1:275$
$\frac{v}{c}=\frac{1}{275}$
$v=\left(\frac{c}{137}\right)\frac{z}{n}$
$\frac{v}{c}=\frac{1}{137}\left(\frac{z}{n}\right)$
for hydrogen z = 1
$\frac{1}{275}=\left(\frac{1}{137}\right)\frac{z}{n}$
$\frac{1}{275}=\frac{1}{137}\left(\frac{1}{n}\right)$
$n=\frac{275}{137}=2$
wave number for $(n+1)=2+1=3$
$\frac{1}{\lambda }=R_H\left(\frac{z^2}{n^2}\right)=R_H\left(\frac{1^2}{3^2}\right)$
$\frac{1}{\lambda }$ = wave number = $\frac{R_H}{9}$