Question

If Rydberg’s constant is \( R \), the longest wavelength of radiation in Paschen series will be \( \frac{\alpha}{7R} \), where \( \alpha = \_\_\_\_\_\_\_ \).

Answer 1

Correct Answer: 144

To determine the value of \( \alpha \) such that the longest wavelength of radiation in the Paschen series is \( \frac{\alpha}{7R} \), we examine the Paschen series for hydrogen, which involves transitions to the \( n=3 \) energy level. The Rydberg formula governs the emitted radiation's wavelength: \[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] For the longest wavelength, the lowest possible higher energy level is \( n_2=4 \) when \( n_1=3 \): \[ \frac{1}{\lambda} = R \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{9} - \frac{1}{16} \right) \] This simplifies to: \[ \frac{1}{9} - \frac{1}{16} = \frac{16-9}{144} = \frac{7}{144} \] Consequently: \[ \frac{1}{\lambda} = \frac{7R}{144} \] This yields: \[ \lambda = \frac{144}{7R} \] By comparing \( \frac{144}{7R} \) to the given expression \( \frac{\alpha}{7R} \), we find \( \alpha = 144 \). This value of 144 is consistent. The longest wavelength of the Paschen series is indeed \( \frac{144}{7R} \), confirming \( \alpha = 144 \).