If longest wavelength for Paschen series in H-atom is α/7R. Find out α

The correct answer is 144.

To find the value of $\alpha$, we use the Rydberg formula for the hydrogen atom:
$\frac{1}{\lambda }=R{Z}^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$
where:
• $\lambda$ is the wavelength of the emitted spectral line.
• $R$ is the Rydberg constant.
• $Z$ is the atomic number of the element (for hydrogen, $Z=1$).
• $n_1$ and $n_2$ are the principal quantum numbers of the lower and higher energy levels, respectively.

For the Paschen series, transitions terminate at the third energy level, meaning:
$n_1=3$

The longest wavelength (${\lambda }_{\text{max}}$) corresponds to the transition with the minimum energy, which occurs from the adjacent energy level:
$n_2=4$

Substituting these values into the Rydberg formula:
$\frac{1}{\lambda }=R{\left(1\right)}^2(\frac{1}{3^2}-\frac{1}{4^2})$
$\frac{1}{\lambda }=R(\frac{1}{9}-\frac{1}{16})$
$\frac{1}{\lambda }=R\left(\frac{16-9}{144}\right)$
$\frac{1}{\lambda }=R\left(\frac{7}{144}\right)$

Taking the reciprocal to solve for $\lambda$:
$\lambda =\frac{144}{7R}$

We are given that the longest wavelength is $\frac{\alpha }{7R}$. Comparing the two expressions:
$\frac{\alpha }{7R}=\frac{144}{7R}$
Therefore, $\alpha =144$.