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

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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 :

(A) $1 : 1$
(B) $2 : 1$ ✓ Correct
(C) $4 : 1$
(D) $1 : 4$

Answer: (B)

Solution

Using Einstein's photoelectric equation:

K E = h ν - ϕ

For the first frequency:

K E_{1} = 3.8 - 0.6 = 3.2

eV

For the second frequency:

K E_{2} = 1.4 - 0.6 = 0.8

eV

Since

K E = \frac{1}{2} m v^{2}

, the ratio of speeds:

\frac{v _{1}}{v _{2}} = \frac{K E _{1}}{K E _{2}} = \frac{3.2}{0.8} = 4 = 2

Answer: (B)

2 : 1

JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Single correct option

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