JEE Main202126 Aug 2021Morning ShiftPhysicsDual Nature of MatterActual
In a photoelectric experiment, ultraviolet light of wavelength 280 nm is used with lithium cathode having work function ϕ = 2 . 5 eV . If the wavelength of incident light is switched to 400 nm , find out the change in the stopping potential. h = 6 . 63 × 10 - 34 J s , c = 3 × 10 8 m s - 1
Options
- A1 . 1   V
- B0 . 6   V
- C1 . 3   V
- D1 . 9   V
Correct answer
C. 1 . 3   V
Step-by-step solution
From the photoelectric equation, e V 0 = h c λ − ϕ Using this equation for the given two cases, e V 1 = h c 280   nm − ϕ e V 2 = h c 400   nm − ϕ On subtracting, e V 1 − V 2 = 1240 120 280 × 400                                       = 1.3   eV ⇒ V 1 − V 2 = 1 . 3   V