NTA Abhyas JEE Main2020PhysicsDual Nature of MatterPractice
If stopping potentials corresponding to wavelength 4000 A ∘ and 4500 A ∘ are 1 . 3 V and 0 . 9 V , respectively, then the work function of the metal is
Options
- A0 . 3   eV
- B1 . 3 eV
- C2 . 3 eV
- D5 eV
Correct answer
C. 2 . 3 eV
Step-by-step solution
eV s = hc λ - ϕ 0 or eV s + ϕ 0 = hc λ or λ = hc eV s + ϕ 0 ⇒ λ 2 λ 1 = eV s 1 + ϕ 0 eV s 2 + ϕ 0 4 5 0 0 4 0 0 0 = 1.3 + ϕ 0 0.9 + ϕ 0 or ϕ 0 = 851.3 - 9 × 0.9 Solving ϕ 0 = 10.4 - 8.1 = 2.3 eV