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11/02/2022 4:56 pm
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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 released. Find the frequency of the emitted photon. (h = 6.6 × 10-34 Js)
(1) 1.45 × 1016 MHz
(2) 0.19 × 1015 MHz
(3) 1.45 × 109 MHz
(4) 9.0 × 1027 MHz
1 Answer
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11/02/2022 5:00 pm
Correct answer: (3) 1.45 × 109 MHz
Explanation:
For every large distance P.E. = 0 and total energy = 2.6 + 0
= 2.6 eV
Finally in first excited state of H atom total energy = -3.4 eV
Loss in total energy = 2.6 - (-3.4) = 6eV
It is emitted as photon
λ = \(\frac{1240}{6}\) = 206 nm
f = \(\frac{3 \times 10^8}{206 \times 10^{-9}}\)
= 1.45 x 1015 MHz
= 1.45 x 109 MHz