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

A $12.5 \mathrm{eV}$ electron beam is used to bombard gaseous hydrogen at room temperature. It will emit:
  1. 2 lines in the Lyman series and 1 line in the Balmar series
  2. 3 lines in the Lyman series
  3. 1 line in the Lyman series and 2 lines in the Balmar series
  4. 3 lines in the Balmer series

Solution

$ \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 $\mathrm{n}_1=2, \mathrm{n}_2=3$

Asked in: JEE Main 2013 (25 Apr Online)

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