JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Modern Physics

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In a Frank-hertz experiment, an electron of energy 5.6 eV passes through mercury vapour and emerges with an energy 0.7 eV. The minimum wavelength of photons emitted by mercury atoms is close to :

(A)1700 nm
(B)2020 nm
(C)220 nm
(D)250 nm✓ Correct

Answer: (D)

Solution

The electron hands over the energy it loses to a mercury atom, so the excitation energy taken up is

Δ E = 5.6 - 0.7 = 4.9

eV. When the atom falls back it releases this as a single photon, and the shortest wavelength corresponds to the whole

4.9

eV coming out at once. Writing

E = \frac{h c}{λ}

with

h c \approx 12400

\cdot

\AA, the wavelength is

λ = \frac{12400}{4.9} \approx 2530

\AA

\approx 253

nm, which is close to

250

nm.

JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Single correct option

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