JEE Main 2026 Physics question
Light source having wavelength 331 nm is used to generate photo-electrons whose stopping potential is 0.2 V. The work function of the used metal in the experiment is α×10⁻¹⁹ J. The value of α is . (h=6.62×10⁻³⁴ J s, e=1.6×10⁻¹⁹ C and c=3×10⁸ m/s)
Answer: (C)
Solution
= hc/λ = 6×10⁻¹⁹ J = 3.75 eV. φ = − eV₀ = 6×10⁻¹⁹ − 1.6×10⁻¹⁹(0.2) = 5.68×10⁻¹⁹ J.