Question

The recoil speed of hydrogen atom after it emits a photon in going from n = 2 state to n = 1 state is nearly [Take $R_{\infty }=1.1\times {10}^{-1}$ and $h=6.63\times {10}^{-34}\text{Js}]$

• A. $1.5{\text{ms}}^{-1}$
• B. $3.3{\text{ms}}^{-1}$
• C. $4.5{\text{ms}}^{-1}$
• D. $6.6{\text{ms}}^{-1}$

Answer 1

The correct option is B $3.3{\text{ms}}^{-1}$

The energy released for the transition is given by
$E=R_{\infty }ch(\frac{1}{(n_1{)}^2}-\frac{1}{(n_2{)}^2})$
$E=R_{\infty }ch(1-\frac{1}{(2{)}^2})$
$E=\frac{3}{4}{R}_{\infty }ch$
The momentum of the photon
$=\frac{E}{c}$
$=\frac{3}{4}{R}_{\infty }h$
From momentum conservation,
$mv=\frac{3}{4}{R}_{\infty }h$
$v=\frac{3}{4}\frac{R_{\infty }h}{m}$
Putting the values
$v=\frac{3\times 1.1\times {10}^7\times 6.63\times {10}^{-34}}{4\times 1.67\times {10}^{-27}}$
$v=3.3\text{m/s}$