Match List I with List II:

List I	List II
A. E = hv$E = hv$	I. de Broglie wavelength
B. Diffraction and Interference	II. Particle nature of light
C. λ = h/p$\lambda = h/p$	III. Wave nature of light
D. Compton effect	IV. Energy of photon

Choose the correct answer from the options given below:

Related Formula

Photoelectric effect condition: E ≥ W₀ λ ≤ λ₀$E \ge W_0 \implies \lambda \le \lambda_0$

Core Logic

Calculate the threshold wavelength (λ₀$\lambda_0$) for the metal. If the incident wavelength λ$\lambda$ is strictly greater than λ₀$\lambda_0$, the energy is insufficient, and the photoelectric effect will NOT occur.

Work function W₀ = 6.6 ~eV = 6.6 × 1.6 × 10⁻¹⁹ ~J$W_0 = 6.6 \mathrm{~eV} = 6.6 \times 1.6 \times 10^{-19} \mathrm{~J}$.

Step 1: Calculate Threshold Wavelength

Step 2: Conclusion

For the photoelectric effect to not occur, the incident wavelength must be larger than 187.5 ~nm$187.5 \mathrm{~nm}$. Looking at the options, only 200 ~nm$200 \mathrm{~nm}$ is greater than 187.5 ~nm$187.5 \mathrm{~nm}$.

Pattern Recognition

Use the highly efficient approximation E(eV) = 1240λ (nm)$E(\text{eV}) = \frac{1240}{\lambda (\text{nm})}$. Threshold λ₀ = (1240)/(6.6) ≈ 187 ~nm$\lambda_0 = \frac{1240}{6.6} \approx 187 \mathrm{~nm}$. Any λ > 187 ~nm$\lambda > 187 \mathrm{~nm}$ fails to eject electrons.

Chapter Mix

Class 12 Physics: Dual Nature of Radiation and Matter