Question

When an electron in a hydrogen atom jumps from the third orbit to the second orbit, it emits a photon of wavelength \( \lambda \). When it jumps from the fourth orbit to the third orbit, the wavelength emitted by the photon will be

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

Hint:

In hydrogen spectrum problems, always compare wavelengths by taking ratios using the Rydberg formula.

Answer 1

The Correct Option is D

Step 1: Use Rydberg formula.
\[ \frac{1}{\lambda} = R\left(\frac{1}{n_2^2} - \frac{1}{n_1^2}\right) \]

Step 2: Transition from 3 to 2.
\[ \frac{1}{\lambda} = R\left(\frac{1}{2^2} - \frac{1}{3^2}\right) = R\left(\frac{5}{36}\right) \]

Step 3: Transition from 4 to 3.
\[ \frac{1}{\lambda'} = R\left(\frac{1}{3^2} - \frac{1}{4^2}\right) = R\left(\frac{7}{144}\right) \]

Step 4: Take ratio.
\[ \frac{\lambda'}{\lambda} = \frac{5/36}{7/144} = \frac{20}{7} \]