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
- A3 λ
- B4 λ
- C3 2 λ
- 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