JEE Main202624 January 2026Evening ShiftPhysicsDual Nature of MatterActual
When a light of a given wavelength falls on a metallic surface the stopping potential for photoelectrons is 3.2 V. If a second light having wavelength twice of first light is used, the stopping potential drops to 0.7 V. The wavelength of first light is _ _ _ _ m. ( h =6.63 10⁻³⁴ ~J . s , e =1.6 10⁻¹⁹ C , c =3 10⁸ ~m / s )
Options
- A2.5 10⁻⁷
- B2.2 10⁻⁸
- C3.1 10⁻⁷
- D2.9 10⁻⁸
Correct answer
A. 2.5 10⁻⁷
Step-by-step solution
Using Einstein's photoelectric equation with stopping potentials: For first light, h ₁ = + eV_ s1 . For second light with ₂ = 2 ₁ (so ₂ = ₁/2 ): h ₁/2 = + eV_ s2 . From given values: h ₁ - 3.2e = h ₁/2 - 0.7e gives h ₁/2 = 2.5e . Thus ₁ = hc 5e = 6.63 10⁻³⁴ 3 10^8 5 1.6 10⁻¹⁹ 2.5 10⁻⁷ m.