A photosensitive surface has work function φ. If photon of energy 3φ falls on this surface, the electron comes out with maximum velocity of 4 × 10⁶ m/s When photon energy is increased to 7φ then maximum velocity of photoelectron will be (a) 4√(3) × 10⁶ m/s (b) 2√(3) × 10⁶ m/s (c) 4√(3) × 10³ m/s (d) 2√(3) × 10³ m/s

Answer: (a). 1/2mv² = E - φ K₁ = 3φ - φ = 2φ, K₂ = 7φ - φ = 6φ v₂ = v₁√(6/2) = 4√(3)×10⁶ m/s So the answer is (a) 4√(3)×10⁶ m/s.

A photosensitive surface has work function φ. If photon of energy 3φ… | MHT CET 2025 (5 May, Shift 2)

Q42MHT CET 2025 · 5 MAY · SHIFT 2Medium

Photoelectric Effect: Einstein's Equation

A photosensitive surface has work function $\phi$ . If photon of energy $3\phi$ falls on this surface, the electron comes out with maximum velocity of $4 \times 10^{6}\text{ m/s}$ When photon energy is increased to $7\phi$ then maximum velocity of photoelectron will be

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Answer: (a)

$\frac{1}{2}mv^2 = E - \phi$
$K_1 = 3\phi - \phi = 2\phi$ , $K_2 = 7\phi - \phi = 6\phi$
$v_2 = v_1\sqrt{\frac{6}{2}} = 4\sqrt{3}\times10^{6}\text{ m/s}$
So the answer is (a) $4\sqrt{3}\times10^{6}$ m/s.

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