Question

Calculate the wavenumber for the shortest wavelength transition in the Balmer series of atomic hydrogen. $\left( R _{H I}=109677 cm ^{-1}\right)$.

Answer 1

SOLUTION : For the Balmer series of the hydrogen atom

$\bar{v}=\frac{1}{\lambda}=R_{H}\left(\frac{1}{2^{2}}-\frac{1}{n^{2}}\right)$

Shortest wavelength means that radiation is of highest energy, hence $n=\infty$.

$\bar{v}=109677 \times \frac{1}{4} cm ^{-1}=27419 \cdot 25 cm ^{-1} \text {. }$