Question

Find the ratio of momentum of photons of 1\textsuperscript{st and 2\textsuperscript{nd} line of Balmer series of hydrogen atom.}

• A. \( \frac{10}{20} \)
• B. \( \frac{11}{27} \)
• C. \( \frac{15}{20} \)
• D. \( \frac{20}{27} \)

Answer 1

The Correct Option is D

Concept: Momentum of a photon is given by \[ p=\frac{h}{\lambda} \] Hence \[ p \propto \frac{1}{\lambda} \] Using Rydberg formula for hydrogen spectrum \[ \frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \] For Balmer series \(n_1=2\). Step 1: First Balmer line \((n_2=3 \rightarrow n_1=2)\).} \[ \frac{1}{\lambda_1} = R\left(\frac{1}{2^2}-\frac{1}{3^2}\right) \] \[ = R\left(\frac{1}{4}-\frac{1}{9}\right) \] \[ = R\left(\frac{5}{36}\right) \]
Step 2: Second Balmer line \((n_2=4 \rightarrow n_1=2)\).} \[ \frac{1}{\lambda_2} = R\left(\frac{1}{4}-\frac{1}{16}\right) \] \[ = R\left(\frac{3}{16}\right) \]
Step 3: Find ratio of momentum.} \[ \frac{p_1}{p_2} = \frac{\lambda_2}{\lambda_1} \] \[ = \frac{\frac{1}{(3R/16)}}{\frac{1}{(5R/36)}} = \frac{16}{3R}\times\frac{5R}{36} \] \[ = \frac{20}{27} \] Final Result \[ \frac{p_1}{p_2}=\frac{20}{27} \]