JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Modern Physics

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Given figure shows few data points in a photo electric effect experiment for a certain metal. The minimum energy for ejection of electron from its surface is: (Plancks constant

h = 6.62 \times 1 0^{- 34}

J.s)

(A)1.93 eV
(B)2.10 eV
(C)2.59 eV
(D)2.27 eV✓ Correct

Answer: (D)

Solution

At the stopping potential the fastest photoelectrons are just halted, and the metal's work function equals the photon energy at the threshold frequency,

ϕ = h ν_{0}

. Reading the threshold frequency from the graph as

ν_{0} = 5.5 \times 1 0^{14}

Hz,

ϕ = (6.62 \times 1 0^{- 34}) (5.5 \times 1 0^{14}) = 3.64 \times 1 0^{- 19}

J. Dividing by

1.6 \times 1 0^{- 19}

J/eV, the minimum energy needed to eject an electron is about

2.27

eV.

JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Single correct option

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