Question

The first emission line in the atomic spectrum of hydrogen in the Balmer series appears at:

• A. $\frac{5R}{36}c{{m}^{-1}}$
• B. $\frac{3R}{4}c{{m}^{-1}}$
• C. $\frac{7R}{144}c{{m}^{-1}}$
• D. $\frac{9R}{400}c{{m}^{-1}}$

Answer 1

Answer: (A)

$\frac{5R}{36}c{{m}^{-1}}$

Answer 2

For Balmer series in the atomic spectrum of hydrogen ${{n}_{1}}=2$ and ${{n}_{2}}=3,$ the $v={{R}_{H}}\left[ \frac{1}{\eta _{1}^{2}}-\frac{1}{n_{2}^{2}} \right]$ $={{R}_{H}}\left[ \frac{1}{{{2}^{2}}}-\frac{1}{{{3}^{2}}} \right]$ $={{R}_{H}}\left[ \frac{1}{4}-\frac{1}{9} \right]$ $={{R}_{H}}\left[ \frac{9-4}{36} \right]$ $={{R}_{H}}\frac{5}{36}$ $=\frac{5R}{36}c{{m}^{-1}}$