JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Modern Physics

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 \times 1 0^{- 34} Js

)

(A) $0.19 \times 1 0^{15} MHz$
(B) $1.45 \times 1 0^{9} MHz$ ✓ Correct
(C) $1.45 \times 1 0^{16} MHz$
(D) $9.0 \times 1 0^{27} MHz$

Answer: (B)

Solution

The free electron carries kinetic energy

2.6

eV, so its total energy is

+ 2.6

eV relative to the ionisation level. It is captured into the first excited state of hydrogen, whose energy is

- 3.4

eV, and the photon removes the energy difference

E = 2.6 - (- 3.4) = 6.0 eV = 9.6 \times 1 0^{- 19} J .

The emitted frequency is

f = \frac{E}{h} = \frac{9.6 \times 1 0 ^{- 19}}{6.6 \times 1 0 ^{- 34}} = 1.45 \times 1 0^{15} Hz = 1.45 \times 1 0^{9} MHz .

JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Single correct option

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JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Modern Physics

Solution

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