Question

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

Answer 1

The formula for wavelength $\left(\lambda \right)$ in Balmer series is:

$\frac{1}{\lambda }=R(\frac{1}{2^2}-\frac{1}{n^2})$

where n=3, 4, 5... and

$R=1.097\times {10}^7{m}^{-1}$ is the Rydberg constant.

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

$\frac{1}{\lambda }=1.097\times {10}^7\times (0.25-0)$

$\frac{1}{\lambda }=0.2743\times {10}^7$

$\lambda =3.6463\times {10}^{-7}$

$=364.63\times {10}^{-9}m$

$=364.63nm$

This wavelength lies in the visible region of the hydrogen spectrum.