Question

The first emission line in the atomic spectrum of hydrogen in the Balmer series appears at:

• A. $\frac{9R}{400}c{m}^{-1}$
• B. $\frac{7R}{144}c{m}^{-1}$
• C. $\frac{3R}{4}c{m}^{-1}$
• D. $\frac{5R}{36}c{m}^{-1}$

Answer 1

The correct option is D $\frac{5R}{36}c{m}^{-1}$

The first emission line in the atomic spectrum of hydrogen in the Balmer series occurs due to the transition from $n_1=2$ to $n_2=3.$
The wavenumber is given by the expression,
$\overline{v}=R(\frac{1}{n_1^2}-\frac{1}{n_2^2})c{m}^{-1}\overline{v}=R(\frac{1}{2^2}-\frac{1}{3^2})c{m}^{-1}\overline{v}=R(\frac{1}{4}-\frac{1}{9})c{m}^{-1}\overline{v}=R\left(\frac{9-4}{4\times 9}\right)c{m}^{-1}\overline{v}=\frac{5R}{36}c{m}^{-1}$
The first emission line in the atomic spectrum of hydrogen in the Balmer series appears at
$\frac{5R}{36}c{m}^{-1}$