Question

If an electron in hydrogen atom jumps from \(3^{\text{rd}}\) orbit to \(2^{\text{nd}}\) orbit it emits a photon of wavelength '\(\lambda\)'. When it emits a photon from \(4^{\text{th}}\) orbit to \(3^{\text{rd}}\) orbit then the corresponding wavelength of emitted photon will be

• A. \(\frac{16}{25}\lambda\)
• B. \(\frac{9}{16}\lambda\)
• C. \(\frac{20}{7}\lambda\)
• D. \(\frac{20}{13}\lambda\)

Hint:

Shortcut: Compare inverse wavelengths using energy differences

Answer 1

The Correct Option is C

Concept: \[ \frac{1}{\lambda} = R \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) \]

Step 1: For transition $3 \to 2$. \[ \frac{1}{\lambda} = R \left(\frac{1}{2^2} - \frac{1}{3^2}\right) = R\left(\frac{1}{4} - \frac{1}{9}\right) = R\left(\frac{5}{36}\right) \]

Step 2: For transition $4 \to 3$. \[ \frac{1}{\lambda'} = R \left(\frac{1}{3^2} - \frac{1}{4^2}\right) = R\left(\frac{1}{9} - \frac{1}{16}\right) = R\left(\frac{7}{144}\right) \]

Step 3: Take ratio. \[ \frac{\lambda'}{\lambda} = \frac{5/36}{7/144} = \frac{5 \times 144}{36 \times 7} = \frac{20}{7} \]

Step 4: Conclusion.
\[ \lambda' = \frac{20}{7}\lambda \] Final Answer: Option (C)