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

Step 1: Understanding the Concept:
The momentum of a photon is inversely proportional to its wavelength. We determine the respective wavelengths of the Balmer series lines using the Rydberg formula and then find their inverse ratio.
Step 2: Key Formula or Approach:
Photon momentum: \(p = \frac{h}{\lambda} \implies p \propto \frac{1}{\lambda}\)
Rydberg formula: \(\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)\)
Step 3: Detailed Explanation:
The Balmer series ends at \(n_1 = 2\). For the \(1^{\text{st}}\) line (\(n_2 = 3 \to n_1 = 2\)): \[ \frac{1}{\lambda_1} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = \frac{5R}{36} \] Thus, \(p_1 \propto \frac{5R}{36}\).
For the \(2^{\text{nd}}\) line (\(n_2 = 4 \to n_1 = 2\)): \[ \frac{1}{\lambda_2} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{4} - \frac{1}{16} \right) = \frac{3R}{16} \] Thus, \(p_2 \propto \frac{3R}{16}\).
Now find the ratio of momenta: \[ \frac{p_1}{p_2} = \frac{1/\lambda_1}{1/\lambda_2} = \frac{5R / 36}{3R / 16} \] \[ \frac{p_1}{p_2} = \left(\frac{5}{36}\right) \times \left(\frac{16}{3}\right) = \frac{80}{108} \] Simplify by dividing the numerator and denominator by 4: \[ \frac{p_1}{p_2} = \frac{20}{27} \] Step 4: Final Answer:
The ratio is \(20/27\).