Two identical photocathodes receive light of frequencies ' n₁ ' and ' n₂ '. If the velocities of the emitted photoelectrons of mass ' m ' are ' V₁ ' and ' V₂ ' respectively, then ( h = Planck's constant) (a) V₁ + V₂ = [2h/m(n₁ + n₂)]1/2 (b) V₁ - V₂ = [2h/m(n₁ - n₂)]1/2 (c) V₁² + V₂² = 2h/m(n₁ + n₂) (d) V₁² - V₂² = 2h/m(n₁ - n₂)

Answer: (d). 1/2mV₁² = hn₁ - φ and 1/2mV₂² = hn₂ - φ (same φ) Subtract: 1/2m(V₁² - V₂²) = h(n₁ - n₂) So the answer is (d) V₁² - V₂² = 2h/m(n₁ - n₂).

Two identical photocathodes receive light of frequencies ' n₁ ' and '… | MHT CET 2024 (9 May, Shift 1)

Q66MHT CET 2024 · 9 MAY · SHIFT 1Medium

Photoelectric Effect: Einstein's Equation

Two identical photocathodes receive light of frequencies ' $n_1$ ' and ' $n_2$ '. If the velocities of the emitted photoelectrons of mass ' $m$ ' are ' $V_1$ ' and ' $V_2$ ' respectively, then ( $h =$ Planck's constant)

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Answer: (d)

$\frac{1}{2}mV_1^2 = hn_1 - \phi$ and $\frac{1}{2}mV_2^2 = hn_2 - \phi$ (same $\phi$ )
Subtract: $\frac{1}{2}m(V_1^2 - V_2^2) = h(n_1 - n_2)$
So the answer is (d) $V_1^2 - V_2^2 = \frac{2h}{m}(n_1 - n_2)$ .

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