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JEE MAIN 2021
31-08-21 S1
Question
A free electron of 2.6 eV energy collides with a $\mathrm{H}_{+}$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. $\left(\mathrm{h}=6.6 \times 10^{-34} \mathrm{Js}\right)$
Select the correct option:
A
$1.45 \times 10^{16} \mathrm{MHz}$
B
$0.19 \times 10^{15} \mathrm{MHz}$
C
$1.45 \times 10^9 \mathrm{MHz}$
D
$9.0 \times 10^{27} \mathrm{MHz}$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
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Question Tags
JEE Main
Physics
Easy
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