JEE Mains · Physics · STD 12 - Atoms
- JEE Mains 2024
- Hard
- Numerical
Hydrogen atom is bombarded with electrons accelerated through a potential different of \(\mathrm{V}\), which causes excitation of hydrogen atoms. If the experiment is being formed at \(\mathrm{T}=0 \mathrm{~K}\). The minimum potential difference needed to observe any Balmer series lines in the emission spectra will be \(\frac{\alpha}{10} \mathrm{~V}\), where \(\alpha=\) _______.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(121\)
Step-by-step Solution
Detailed explanation
For minimum potential difference electron has to make transition from \(n=3\) to \(n=2\) state but first electron has to reach to \(\mathrm{n}=3\) state from ground state. So, energy of bombarding electron should be equal to energy difference of \(n=3\) and \(n=1\) state.
\( \Delta \mathrm{E}=13.6\left[1-\frac{1}{3^2}\right] \mathrm{e}=\mathrm{eV} \)
\( \frac{13.6 \times 8}{9}=\mathrm{V} \)
\( \mathrm{V}=12.09 \mathrm{~V} \approx 12.1 \mathrm{~V}\)
So, \(\alpha=121\)
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