Q84 / 120 · VITEEE 2007 · Physics
Assuming $\mathrm{f}$ to be the frequency of first line in Balmer series, the frequency of the immediate next (i.e. second) line is
- A$0.50 \mathrm{f}$
- Correct answerB$1.35 \mathrm{f}$
- C$2.05 \mathrm{f}$
- D$2.70 \mathrm{f}$
Solution
Balmer series is given for $\mathrm{n}_{1}=2$ and $\mathrm{n}_{2}=3,4, \ldots$ $v=\operatorname{Rc}\left[\frac{1}{2^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right]$
For Ist line in spectrum $n_{2}=3$
$\begin{array}{l}
v_{1}=\operatorname{Rc}\left[\frac{1}{2^{2}}-\frac{1}{n_{2}^{2}}\right]=\operatorname{Rc}\left[\frac{1}{4}-\frac{1}{9}\right]=\mathrm{f} \\
\Rightarrow f=\frac{5}{36} R c \Rightarrow R c=\frac{36}{5} \mathrm{f}
\end{array}$
For second line $\mathrm{n}_{2}=4$
$\begin{aligned}
v_{2}=\operatorname{Rc}\left[\frac{1}{2^{2}}-\frac{1}{4^{2}}\right] &=\operatorname{Rc}\left[\frac{1}{4}-\frac{1}{16}\right] \\
&=\operatorname{Rc}\left[\frac{4-1}{16}\right]
\end{aligned}$
$v_{2}=\frac{3}{16} \mathrm{Rc}=\frac{3}{16} \times \frac{36}{5} \mathrm{f}=1.35 \mathrm{f}$
Hence, the correct answer is option: 'B'.