For Balmer series
$\frac{1}{\lambda} =R \left[\frac{1}{2^{2} } -\frac{1}{n^{2}}\right]$ where $n = 3, 4 , 5 .....(i)$
By putting n = ∞ in equation (i), we obtain the series limit of the Balmer series
$\frac{1}{\lambda} =R \left[\frac{1}{2^{2} } -\frac{1}{∞^{2}}\right]$
or $\lambda $ = 364.5 nm.
This is the shortest wavelength in the series
Energy of photon $E = \frac{hc}{\lambda}$
Since $E = \propto \frac{1}{\lambda}$
For shortest wavelength, energy of photon is the most.