Question

What is the wave number of $4^{th}$ line in Balmer series of hydrogen spectrum?

$[R=1,09,677c{m}^{-1}]$

• A. $24,630c{m}^{-1}$
• B. $24,360c{m}^{-1}$
• C. $24,730c{m}^{-1}$
• D. $24,372c{m}^{-1}$

Answer 1

The correct option is D $24,372c{m}^{-1}$

Rydberg equation is :

$\nu =\frac{1}{\lambda }=R(\frac{1}{n_f^2}-\frac{1}{n_i^2})$

$n_i$ and $n_f$ are initial and final states.

$\nu$ is wave number.

For $4th$ line in Balmer series, initial and final states are 6 and 2, respectively.

$\nu =109677\times (\frac{1}{4}-\frac{1}{36})=24372.66c{m}^{-1}$