JEE MAIN - Physics (2024 - 29th January Morning Shift - No. 24)
When a hydrogen atom going from $$n=2$$ to $$n=1$$ emits a photon, its recoil speed is $$\frac{x}{5} \mathrm{~m} / \mathrm{s}$$. Where $$x=$$ ________. (Use, mass of hydrogen atom $$=1.6 \times 10^{-27} \mathrm{~kg}$$)
Answer
17
Explanation
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$$\begin{aligned} & \Delta \mathbf{E}=\mathbf{1 0 . 2} \mathrm{eV} \\ & \text { Recoil speed }(\mathrm{v})=\frac{\Delta \mathrm{E}}{\mathrm{mc}} \\ & =\frac{10.2 \mathrm{eV}}{1.6 \times 10^{-27} \times 3 \times 10^8} \\ & =\frac{10.2 \times 1.6 \times 10^{-19}}{1.6 \times 10^{-27} \times 3 \times 10^8} \\ & \mathrm{v}=3.4 \mathrm{~m} / \mathrm{s}=\frac{17}{5} \mathrm{~m} / \mathrm{s} \end{aligned}$$
Therefore, $$x=17$$
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