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JEE Mains · Physics · STD 12 - Atoms
- JEE Mains 2013
- Hard
- MCQ
A \(12.5\, eV\) electron beam is used to bombard gaseous hydrogen at room temperature. It will emit
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(2\) lines in the Lyman series and \(1\) line in the Balmar series
Step-by-step Solution
Detailed explanation
\(\mathrm{E}=\frac{\mathrm{hc}}{\lambda} \Rightarrow \lambda=\frac{\mathrm{hc}}{\mathrm{E}}=\frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{12.5 \times 1.6 \times 10^{-19}}\)
\(=993 \mathrm{A}^{\circ}\)
\(\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right)\)
(where Rydberg constant, \(\mathrm{R}=1.097 \times 10^{7}\) )
\(\frac{1}{993 \times 10^{-10}}=1.097 \times 10^{7}\left(\frac{1}{1^{2}}-\frac{1}{n_{2}^{2}}\right)\)
Solving we get \(n_{2}=3\)
Spectral lines
Total number of spectral lines \(=3\)
Two lines in Lyman series for \(n_{1}=1, n_{2}=2\) and \(\mathrm{n}_{1}=1, \mathrm{n}_{2}=3\) and one in Balmer series for \(n_{1}=2, n_{2}=3\)
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