Thank you for reporting, we will resolve it shortly
Q.
The wavelength limit present in the Pfund series is
$(R = 1.097 \times 10^7\,m^{-1})$
Atoms
Solution:
The wavelength for Pfund series is given by
$\frac{1}{\lambda} = R\left[\frac{1}{5^{2}} -\frac{1}{n^{2}}\right] $
for series limit, $ n =\infty $
$ \therefore \frac{1}{\lambda} = R \left[\frac{1}{25} -\frac{1}{\infty^{2}}\right] = \frac{R}{25} $
$\therefore \lambda = \frac{25}{R} $
$= \frac{25}{1.097 \times10^{7}} $
$ = 2278 \,nm$