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 $c_{2}=\frac{4}{R_{H}}$