Question

A hydrogen atom emits a photon from $\mathrm{n}=5$ state to $\mathrm{n}=1$. What will be its recoiling speed?

• A. $4.17\mathrm{m}/s$
• B. $4.34\mathrm{m}/\mathrm{s}$
• C.
• D.

Answer 1

The correct option is A

$4.17\mathrm{m}/s$

The explanation for the correct options:

A. $4.17\mathrm{m}/\mathrm{s}$

Step 1:

$∆\mathrm{E}$ is the energy released when a photon goes from one level to another level.

$\begin{array}{rcl}\mathrm{n}& =& 5\mathrm{to}1\\ ∆\mathrm{E}& =& 13.6-0.54\\ & =& 13.06\mathrm{eV}\end{array}$

Step 2:

Momentum$\mathrm{P}$=$\frac{∆\mathrm{E}}{\mathrm{C}}$

Where $∆\mathrm{E}$is the change in energy,$\mathrm{C}$ is the speed of light.

${\mathrm{P}}_{\mathrm{i}}={\mathrm{P}}_{\mathrm{f}}$

$0=\frac{\mathrm{h}}{\mathrm{\lambda }}-\mathrm{Mv}$

$\begin{array}{rcl}{\mathrm{V}}_{\mathrm{recoil}}& =& \frac{\mathrm{h}}{\mathrm{\lambda M}}\cdots \left(\mathrm{i}\right)\\ ∆\mathrm{E}& =& \frac{\mathrm{hc}}{\mathrm{\lambda }}=\frac{\mathrm{hc}}{\mathrm{\lambda M}}\times \mathrm{M}\end{array}$

where M is the mass of the proton.

Step3:

$\begin{array}{rcl}{\mathrm{V}}_{\mathrm{recoil}}& =& \frac{∆\mathrm{E}}{\mathrm{Mc}}\\ & & \\ & =& \frac{13.06\times 1.06\times {10}^{-19}}{1.67\times {10}^{-27}\times 3\times {10}^8}\\ & =& 4.17\mathrm{m}/\mathrm{s}\end{array}$

Hence option (A) is correct.,$4.17\mathrm{m}/\mathrm{s}$ will be its recoiling speed.