Question

Calculate the wavelength of the first line in the Balmer series of hydrogen spectrum.

• A. 486 nm
• B. 911.7 $\mathring{A}$
• C. 656.5 nm
• D. 434 nm

Answer 1

The correct option is C

656.5 nm

From line spectrum of hydrogen, we know that
$\overline{\nu }=R_H[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
$=109677[\frac{1}{(2{)}^2}-\frac{1}{(3{)}^2}]c{m}^{-1}$

$=109677\times \frac{5}{36}c{m}^{-1}$
$=1523.9c{m}^{-1}$
Or $\lambda =\frac{1}{\overline{\nu }}=\frac{1}{15,232.9}cm=6.565\times {10}^{-1}cm=656.5nm$