Question

Calculate the shortest wavelength in the Balmer series of hydrogen atom. In which region (infra-red, visible, ultraviolet) of hydrogen spectrum does this wavelength lie?

Answer 1

Wavelength in the Balmer series of H-atom:

$\frac{1}{\lambda }=1.097\times {10}^7\times (\frac{1}{n_1^2}-\frac{1}{n_2^2})$ $m^{-1}$

For shortest wavelength in the Balmer series: $n_1=2$ $n_2=\infty$

$\therefore$ $\frac{1}{{\lambda }_{min}}=1.097\times {10}^7\times (\frac{1}{2^2}-\frac{1}{\infty })$$m^{-1}$

${\lambda }_{min}=\frac{4}{1.097\times {10}^7}m$ $=3.646\times {10}^{-7}m=364.6$ $nm$

This wavelength lies in the ultraviolet region.