Question

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

Answer 1

The Paschen series corresponds to transitions to $n = 3$. The longest wavelength corresponds to the transition between $n = 4$ and $n = 3$. The inverse wavelength is given by:

$\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$

For $n_1 = 3$ and $n_2 = 4$, and taking $Z = 1$:

$\frac{1}{\lambda} = R \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{9} - \frac{1}{16} \right)$

$\frac{1}{\lambda} = R \left( \frac{16 - 9}{144} \right) = \frac{7R}{144}$

Thus:

$\alpha = 144$
The Correct answer is: 144