Physics · Atomic Physics

JEE Main 2024 — 29 January, Shift 2 — Question 51

Hydrgen atom is bombarded with electrons accelerated through a potential different of V, which causes excitation of hydrogen atom. If the experiment is being formed at T = 0 K. The minimum potential different needed to observe any Balmer series lines in the emission spectra will be $\frac{\alpha }{10}V$ , where $\alpha$ =___

Answer: 121

Numerical answer — enter this value.

Step-by-step solution

For minimum potential difference electron has to make transition from $\mathrm{n}=3$ to $\mathrm{n}=2$

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 $\mathrm{n}=3$ and $\mathrm{n}=1$

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, $\alpha=121$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 29 January, Shift 2
Subject: Physics
Chapter: Atomic Physics
Topic: Hydrogen Spectrum

Answer: 121

Step-by-step solution

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