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Match List-I with List-II regarding a photoelectric experiment: List-I List-II (A) Intensity of incident light is doubled at constant frequency (I) Stopping potential decreases (B) Frequency of incident light is increased keeping photon emission rate constant (II) Saturation photocurrent becomes one-fourth (C) Distance of a point source of light is doubled (III) Saturation photocurrent is doubled (D) Photosensitive m

Options

  1. A(A)-(III), (B)-(II), (C)-(IV), (D)-(I)
  2. B(A)-(IV), (B)-(III), (C)-(II), (D)-(I)
  3. C(A)-(III), (B)-(IV), (C)-(I), (D)-(II)
  4. D(A)-(III), (B)-(IV), (C)-(II), (D)-(I)

Correct answer

D. (A)-(III), (B)-(IV), (C)-(II), (D)-(I)

Step-by-step solution

(A) Saturation photocurrent is directly proportional to the intensity of incident light (for a given frequency). Doubling the intensity doubles the saturation current. (A III) (B) According to Einstein's photoelectric equation, eV₀ = h - . Increasing the frequency increases the stopping potential V₀ . (B IV) (C) For a point source, intensity I 1 r^2 . Doubling the distance r makes the intensity one-fourth, so the saturation photocurrent becomes one-fourth. (C II) (D) From eV₀ = h - , if the work function is higher

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