Question 23 of 239

A $12.5 \mathrm{eV}$ electron beam is used to bombard gaseous hydrogen at room temperature. It will emit:

A2 lines in the Lyman series and 1 line in the Balmar seriesB3 lines in the Lyman seriesC1 line in the Lyman series and 2 lines in the Balmar seriesD3 lines in the Balmer series

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Answer: (A) 2 lines in the Lyman series and 1 line in the Balmar series

$$\begin{aligned} &\mathrm{E}=\frac{\mathrm{hc}}{\lambda} \Rightarrow \lambda=\frac{\mathrm{hc}}{\mathrm{E}}=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{12.5 \times 1.6 \times 10^{-19}} \\ &=993 \mathrm{~A}^{\circ} \\ &\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right) \\ & \end{aligned}$$
(where Rydberg constant , $\mathrm{R}=1.097 \times 10^7$ )
or, $\frac{1}{993 \times 10^{-10}}=1.097 \times 10^7\left(\frac{1}{1^2}-\frac{1}{\mathrm{n}_2^2}\right)$
Solving we get $\mathrm{n}_2=3$
Spectral lines
Total number of spectral lines $=3$
Two lines in Lyman series for $\mathrm{n}_1=1, \mathrm{n}_2=2$ and $\mathrm{n}_1=1, \mathrm{n}_2=3$ and one in Balmer series for $\mathrm{n}_1=2, \mathrm{n}_2=3$

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