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Physics · Atomic Physics

JEE Main 2024 — 29 January, Shift 1 — Question 49

When a hydrogen atom going from n=2n=2 to n=1n=1 emits a photon, its recoil speed is X5 m/s\frac{\mathrm{X}}{5} \mathrm{~m} / \mathrm{s}. Where x=\mathrm{x}= (Use : mass of hydrogen atom =1.6×10−27 kg)\left.=1.6 \times 10^{-27} \mathrm{~kg}\right)

Answer: 17

Numerical answer — enter this value.

Step-by-step solution

figure

ΔE=10.2eV\Delta E=10.2 \mathrm{eV}

Recoil⁡speed⁡(v)=ΔEmc\operatorname{Recoil} \operatorname{speed}(\mathbf{v})=\frac{\Delta \mathrm{E}}{\mathrm{mc}}

=10.2eV1.6×10−27×3×108=\frac{10.2 \mathrm{eV}}{1.6 \times 10^{-27} \times 3 \times 10^{8}} 4=10.2×1.6×10−191.6×10−27×3×108=\frac{10.2 \times 1.6 \times 10^{-19}}{1.6 \times 10^{-27} \times 3 \times 10^{8}}

v=3.4 m/s=175 m/s\mathrm{v}=3.4 \mathrm{~m} / \mathrm{s}=\frac{17}{5} \mathrm{~m} / \mathrm{s}

Therefore, x=17\mathrm{x}=17

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
29 January, Shift 1
Subject
Physics
Chapter
Atomic Physics
Topic
Rutherford's and Bohr's Model of Atom

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