Question

Calculate the wave number for the shortest wavelength transition in the Balmer series of atomic hydrogen.
Rydberg constant = 109677 $c{m}^{-1}$

• A. 109677.84 cm${}^{-1}$
• B. 15232.92 cm${}^{-1}$
• C. 27419.25 cm${}^{-1}$
• D. 20564.44 cm${}^{-1}$

Answer 1

The correct option is C 27419.25 cm${}^{-1}$

The shortest wavelength transition in the Balmer series corresponds to the transition: $n=2\to n=\infty$
The shortest wavelength requires the highest energy for an electron to jump from n = 2 to $n=\infty$
So, $n_1=2and{n}_2=\infty$
Using the formula:
$\overline{v}=R_H(\frac{1}{n_1^2}-\frac{1}{n_2^2})$ where $R_H$ = Rydberg constant
$\overline{v}=\left(109677c{m}^{-1}\right)(\frac{1}{2^2}-\frac{1}{{\infty }_2})Onsolving,\overline{v}=27419.25c{m}^{-1}$