JEE Mains · Physics · STD 12 - Atoms
- JEE Mains 2024
- Hard
- Numerical
When a hydrogen atom going from \(\mathrm{n}=2\) to \(\mathrm{n}=1\) emits a photon, its recoil speed is \(\frac{\mathrm{x}}{5} \mathrm{~m} / \mathrm{s}\). Where \(\mathrm{x}=\) _______. (Use : mass of hydrogen atom\(\left.=1.6 \times 10^{-27 /} \ \mathrm{kg}\right)\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(19\)
Step-by-step Solution
Detailed explanation
\( \Delta E=10.2 \mathrm{eV} \)
\( \text { Recoil speed(v) }=\frac{\Delta \mathrm{E}}{\mathrm{mc}} \)
\( =\frac{10.2 \mathrm{eV}}{1.6 \times 10^{-27} \times 3 \times 10^8} \)
\( =\frac{10.2 \times 1.6 \times 10^{-19}}{1.6 \times 10^{-27} \times 3 \times 10^8} \)
\( \mathrm{v}=3.4 \mathrm{~m} / \mathrm{s}=\frac{17}{5} \mathrm{~m} / \mathrm{s}\)
Therefore, \(x=17\)
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