The threshold frequency of a metal is f₀. When the light of frequency 2f₀ is incident on the metal plate, the maximum velocity of photoelectrons is v₁. When the frequency of incident radiation is increased to 5f₀, the maximum velocity of photoelectrons emitted is v₂. The ratio of v₁ to v₂ is: (a) (v₁)/(v₂) = 1/2 (b) (v₁)/(v₂) = 1/16 (c) (v₁)/(v₂) = 1/4 (d) (v₁)/(v₂) = 1/8

Answer: (a). 1/2mv² = hν - hν₀ At 2ν₀: K₁ = hν₀; at 5ν₀: K₂ = 4hν₀ (v₁)/(v₂) = √(1/4) = 1/2 So the answer is (a) (v₁)/(v₂) = 1/2.

The threshold frequency of a metal is f₀. When the light of frequency… | JEE Main 2023 (1 Feb, Shift 2)

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Photoelectric Effect: Einstein's Equation

The threshold frequency of a metal is $f_0$ . When the light of frequency $2f_0$ is incident on the metal plate, the maximum velocity of photoelectrons is $v_1$ . When the frequency of incident radiation is increased to $5f_0$ , the maximum velocity of photoelectrons emitted is $v_2$ . The ratio of $v_1$ to $v_2$ is:

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Answer: (a)

$\frac{1}{2}mv^2 = h\nu - h\nu_0$
At $2\nu_0$ : $K_1 = h\nu_0$ ; at $5\nu_0$ : $K_2 = 4h\nu_0$
$\frac{v_1}{v_2} = \sqrt{\frac{1}{4}} = \frac{1}{2}$
So the answer is (a) $\frac{v_1}{v_2} = \frac{1}{2}$ .

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