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

Answer: (A)

$\lambda =c_2\left[\frac{n^2}{n^2-2^2}\right]$
or $\frac{1}{\lambda }=\frac{1}{c_2}\left[\frac{n^2-2^2}{n^2}\right]=\frac{1}{c_2}\left[1-\frac{2^2}{n^2}\right]$
$=\frac{2^2}{c_2}\left[\frac{1}{2^2}-\frac{1}{n^2}\right]$
So, $R_H=\frac{2^2}{c_2}$ or $c_2=\frac{4}{R_H}$