Paschen series transition

In the Paschen series, all spectral lines correspond to transitions ending at: $n_f = 3$

Condition for longest wavelength

The longest wavelength corresponds to the smallest energy difference in the series.

So we take the transition from the nearest higher level: $n_i = 4 \to n_f = 3$

Use Rydberg formula

$\frac{1}{\lambda} = R\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$

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

Simplify

$\frac{1}{9} - \frac{1}{16} = \frac{16-9}{144} = \frac{7}{144}$

Hence, $\frac{1}{\lambda} = R\cdot \frac{7}{144}$

Therefore, $\lambda = \frac{144}{7R}$

Compare with given form

Given: $\lambda = \frac{\alpha}{7R}$

So, $\alpha = 144$

Comparison with stored answer

Stored correct answer = $144$

This matches our derived answer.