Question

The wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen is $1.523\times {10}^6{m}^{-1}$

Enter 1 if the statement is True or 0 if False.

Answer 1

The wave number is obtained by the following expression
$\overline{\nu }=R(\frac{1}{n_1^2}-\frac{1}{n_2^2})=1.097\times {10}^7/m\times (\frac{1}{2^2}-\frac{1}{3^2})=1.523\times {10}^6/m$