Atoms & Nuclei

JEE Main level, single correct, medium

127 of 239

Question 127 of 239

A free electron of 2.6eV2.6eV energy collides with a H+{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. (h=6.6×10−34Js)\left(h=6.6\times {10}^{-34}Js\right)

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Answer: (B) 1.45×109MHz1.45\times {10}^{9}MHz

By Energy conservation
(K.E.)e=T.E.H++EPhoton{\left(K.E.\right)}_{e}={T.E.}_{{H}^{+}}+{E}_{Photon}
2.6=−13.64+hf2.6=\dfrac{-13.6}{4}+hf
hf=6eVhf=6eV
f=6×1.6×10−196.626×10−34f=\dfrac{6\times 1.6\times {10}^{-19}}{6.626\times {10}^{-34}}
f=1.45×1015Hz=1.45×109MHzf=1.45\times {10}^{15}Hz=1.45\times {10}^{9}MHz

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