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JEE Main202311 Apr 2023Morning ShiftPhysicsDual Nature of MatterActual

A metallic surface is illuminated with radiation of wavelength λ , the stopping potential is V 0 . If the same surface is illuminated with radiation of wavelength 2 λ , the stopping potential becomes V 0 4 . The threshold wavelength for this metallic surface will be

Options

  1. A3 λ
  2. B4 λ
  3. C3 2 λ
  4. Dλ 4

Correct answer

A. 3 λ

Step-by-step solution

Let the threshold frequency be λ 0 . By the equation of photoelectric effect, for wavelength λ , e V 0 = h c 1 λ - 1 λ 0       . . . ( i ) For the wavelength 2 λ , e V 0 4 = h c 1 2 λ - 1 λ 0       . . . ( i i ) Dividing (i) by (ii) 4 = h c ( λ 0 - λ λ λ 0 ) h c λ 0 - 2 λ 2 λ λ 0 ⇒ 4 = 2 ( λ 0 - λ ) λ 0 - 2 λ ⇒ 2 λ 0 = 6 λ ⇒ λ 0 = 3 λ

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