JEE Main 2022 · 24 Jun · Shift 2 · Physics Q48

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PhysicsSingle correct+4 / -1Question 48 of 90

The light of two different frequencies whose photons have energies 3.8 eV and 1.4 eV respectively, illuminate a metallic surface whose work function is 0.6 eV successively. The ratio of maximum speeds of emitted electrons for the two frequencies respectivly will be :

1 : 1

2 : 1

4 : 1

1 : 4

Answer and worked solution

Official answer: B

Kinetic energies. $3.8-0.6=3.2$ eV and $1.4-0.6=0.8$ eV.

Speeds. $v\propto\sqrt K$ , so $\dfrac{v_1}{v_2}=\sqrt{\dfrac{3.2}{0.8}}=2$ , i.e. $2:1$ .

(B)

Topics:Photoelectric effect

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JEE Main 2022 · 24 Jun · Shift 2, Physics question 48