Question

If $\lambda =c_2\left[\frac{n^2}{n^2-2^2}\right]$ for Balmer series, what is the value of $c_2$?

• A. $\frac{4}{R_H}$
• B. $\frac{2}{R_H}$
• C. $2R_H$
• D. $4R_H$

Answer 1

The correct option is A $\frac{4}{R_H}$

As we know,
for balmer series
$1/\lambda =R_H(1/2^2-1/n^2)$
and here given is
$\lambda =c_2\left[\frac{n^2}{n^2-2^2}\right]$
Or $\frac{1}{\lambda }=\frac{1}{c_2}\left[\frac{n^2-2^2}{n^2}\right]=\frac{1}{c_2}[1-\frac{2^2}{n^2}]$
$=\frac{2^2}{c_2}[\frac{1}{2^2}-\frac{1}{n^2}]$
by comparing
$\therefore R_H=\frac{2^2}{c_2}$
$c_2=\frac{4}{R_H}$