Physics · 2021
JEE Main · 31 August 2021, Shift 2 · Q2
A free electron of 2.6 eV energy collides with a H^+ ion. This results in the formation of a hydrogen atom in the first excited state and a photon is…
A free electron of $\displaystyle 2.6$ eV energy collides with a $\displaystyle \mathrm{H}^{+}$ion. This results in the formation of a hydrogen atom in the first excited state and a photon is released. Find the frequency of the emitted photon. ( $\displaystyle \mathrm{h}=6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s}$ )
Official answer
(3)
$\displaystyle 1.45 \times 10^9 \mathrm{MHz}$
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JEE Main 2021 Physics question, with the answer from NTA’s final answer key. Where our answers come from.