A) \[\lambda =\frac{16}{3R}\]
B) \[\lambda =\frac{36}{5R}\]
C) \[\lambda =\frac{4}{3R}\]
D) None of the above
Correct Answer: A
Solution :
| [a] For Balmer series \[\frac{1}{\lambda }=R\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{n}^{2}}} \right)\] \[\frac{1}{\lambda }=R\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{4}^{2}}} where n = 3, 4, 5 |
| For second line n = 4 |
| So \[\frac{1}{\lambda }=R\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{4}^{2}}} \right)=\frac{3}{16}R\Rightarrow \lambda =\frac{16}{3R}\] |
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