Related Formula
(hc)/(λ) = φ + Kmax$$\frac{hc}{\lambda} = \phi + K_{\text{max}}$$
Kmax = eV₀$$K_{\text{max}} = eV_0$$
φ = (hc)/(λ₀)$$\phi = \frac{hc}{\lambda_0}$$
Core Logic
Using Einstein's Photoelectric equation for the two cases:
Case 1 (Wavelength λ$\lambda$, Stopping Potential 8 V$8\mathrm{\,V}$):
(hc)/(λ) = (hc)/(λ₀) + 8e (i)$$\frac{hc}{\lambda} = \frac{hc}{\lambda_0} + 8e \quad \dots \dots (\text{i})$$
Case 2 (Wavelength 3λ$3\lambda$, Stopping Potential 2 V$2\mathrm{\,V}$):
(hc)/(3λ) = (hc)/(λ₀) + 2e (ii)$$\frac{hc}{3\lambda} = \frac{hc}{\lambda_0} + 2e \quad \dots \dots (\text{ii})$$
Step 2: Solving the Equations
Multiply equation (ii) by 4 to eliminate e$e$:
(4hc)/(3λ) = (4hc)/(λ₀) + 8e$$\frac{4hc}{3\lambda} = \frac{4hc}{\lambda_0} + 8e$$
Equating this to equation (i):
(hc)/(λ) - (hc)/(λ₀) = (4hc)/(3λ) - (4hc)/(λ₀)$$\frac{hc}{\lambda} - \frac{hc}{\lambda_0} = \frac{4hc}{3\lambda} - \frac{4hc}{\lambda_0}$$
Divide entirely by hc$hc$:
(1)/(λ) - (1)/(λ₀) = (4)/(3λ) - (4)/(λ₀)$$\frac{1}{\lambda} - \frac{1}{\lambda_0} = \frac{4}{3\lambda} - \frac{4}{\lambda_0}$$
(4)/(λ₀) - (1)/(λ₀) = (4)/(3λ) - (1)/(λ)$$\frac{4}{\lambda_0} - \frac{1}{\lambda_0} = \frac{4}{3\lambda} - \frac{1}{\lambda}$$
(3)/(λ₀) = (4 - 3)/(3λ)$$\frac{3}{\lambda_0} = \frac{4 - 3}{3\lambda}$$
(3)/(λ₀) = (1)/(3λ)$$\frac{3}{\lambda_0} = \frac{1}{3\lambda}$$
λ₀ = 9λ$$\lambda_0 = 9\lambda$$
Chapter Mix
Class 12 Physics: Dual Nature Of Radiation And Matter