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 i $s_________m.$
(h = 6.63 × 10⁻³⁴ J·s, e = 1.6 × 10⁻¹⁹ C, c = 3 × 10⁸ m/s)

Question

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 i $s_________m.$

(h = 6.63 × 10⁻³⁴ J·s, e = 1.6 × 10⁻¹⁹ C, c = 3 × 10⁸ m/s)

G1 Physics · Accepted Answer

For the first light, e(3.2) = hc/λ - φ. (1) For the second light of wavelength 2λ, e(0.7) = hc/(2λ) - φ. (2) Subtract equation (2) from equation (1): e(3.2-0.7) = hc/λ - hc/(2λ). Therefore, 2.5e = hc/(2λ). Hence, λ = hc/(5e). Substitute: h = 6.63×10⁻³⁴ J·s, c = 3×10⁸ m/s, e = 1.6×10⁻¹⁹ C. Thus, λ = (6.63×10⁻³⁴×3×10⁸)/ (5×1.6×10⁻¹⁹) ≈ 2.49×10⁻⁷ m. Therefore, λ ≈ 2.5×10⁻⁷ m. Correct option: D.