Question

The ratio of the shortest and the longest wavelengths observed in all the five spectral series of hydrogen spectrum is

• A. 5 : 36
• B. 11 : 300
• C. 11 : 600
• D. 11 : 900

Hint:

Use $\dfrac{1}{\lambda} = R \left( \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \right)$ and compare across series for minimum and maximum wavelengths.

Answer 1

The Correct Option is D

Hydrogen spectrum: $\dfrac{1}{\lambda} = R \left( \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \right)$
Shortest wavelength → $n_2 \rightarrow \infty$, $\dfrac{1}{\lambda_{min}} = R \cdot \dfrac{1}{n_1^2}$
Longest wavelength → $n_2 = n_1 + 1$, $\dfrac{1}{\lambda_{max}} = R \left( \dfrac{1}{n_1^2} - \dfrac{1}{(n_1+1)^2} \right)$
For the Lyman series: $\lambda_{min} = \dfrac{1}{R},\ \lambda_{max} = \dfrac{1}{R(1 - 1/4)} = \dfrac{1}{R \cdot 3/4} = \dfrac{4}{3R}$
Ratio for all 5 series combines to maximum across Lyman to Pfund, etc., resulting in $\dfrac{\lambda_{min}}{\lambda_{max}} = \dfrac{11}{900}$