Solution
Related Formula
(hc)/(λ) = φ + Kmax Kmax = eV₀ φ = (hc)/(λ₀)Core Logic
Using Einstein's Photoelectric equation for the two cases:
Case 1 (Wavelength λ, Stopping Potential 8 V):
(hc)/(λ) = (hc)/(λ₀) + 8e (i)Case 2 (Wavelength 3λ, Stopping Potential 2 V):
(hc)/(3λ) = (hc)/(λ₀) + 2e (ii)Step 2: Solving the Equations
Multiply equation (ii) by 4 to eliminate e:
(4hc)/(3λ) = (4hc)/(λ₀) + 8eEquating this to equation (i):
(hc)/(λ) - (hc)/(λ₀) = (4hc)/(3λ) - (4hc)/(λ₀)Divide entirely by hc:
(1)/(λ) - (1)/(λ₀) = (4)/(3λ) - (4)/(λ₀) (4)/(λ₀) - (1)/(λ₀) = (4)/(3λ) - (1)/(λ) (3)/(λ₀) = (4 - 3)/(3λ) (3)/(λ₀) = (1)/(3λ) λ₀ = 9λChapter Mix
Class 12 Physics: Dual Nature Of Radiation And Matter