Question

The wave number for the shortest wavelength transition in the Balmer series of atomic hydrogen is :

• A. $8225714m^{-1}$
• B. $1523693m^{-1}$
• C. $5333c{m}^{-1}$
• D. $27419c{m}^{-1}$

Answer 1

The correct option is D $27419c{m}^{-1}$

The shortest wavelength transition corresponds to $n_2=\infty$ to $n_1=2$ transition.
$\overline{V}=109678[\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2}]{cm}^{-1}$
$=109678[\frac{1}{2^2}-\frac{1}{\infty }]=\frac{109678}{4}=27419{cm}^{-1}$