Correct Answer: (A) **Step-by-step derivation:** 1. **Recall Einstein's Photoelectric Equation:** The photoelectric effect describes the emission of electrons when light shines on a material. Einstein's photoelectric equation relates the energy of the incident photon, the work function of the material, and the maximum kinetic energy of the emitted photoelectrons. It is given by: \[ E = \phi + K_{\text{max}} \] where: * $E$ is the energy of the incident photon. * $\phi$ is the work function of the metal (minimum energy required to eject an electron). * $K_{\text{max}}$ is the maximum kinetic energy of the emitted photoelectrons. 2. **Express Photon Energy in terms of Wavelength:** The energy of a photon can also be expressed in terms of Planck's constant ($h$), the speed of light ($c$), and the wavelength ($\lambda$) as: \[ E = \frac{hc}{\lambda} \] 3. **Apply to the Initial Condition:** For the initial condition, light of wavelength $\lambda$ results in a maximum kinetic energy $K_{\text{max}}$. Substituting the expression for photon energy into Einstein's equation, we get: \[ \frac{hc}{\lambda} = \phi + K_{\text{max}} \quad \text{(Equation 1)} \] 4. **Determine the Work Function ($\phi$):** From Equation 1, we can express the work function $\phi$ as: \[ \phi = \frac{hc}{\lambda} - K_{\text{max}} \quad \text{(Equation 2)} \] The work function $\phi$ is a property of the photosensitive material and remains constant regardless of the incident light's wavelength. 5. **Apply to the New Condition:** Now, the wavelength of the incident light is changed to $\lambda' = \lambda/3$. Let the new maximum kinetic energy be $K_{\text{max}}'$. Using Einstein's photoelectric equation for this new condition: \[ \frac{hc}{\lambda'} = \phi + K_{\text{max}}' \] Substitute $\lambda' = \lambda/3$: \[ \frac{hc}{(\lambda/3)} = \phi + K_{\text{max}}' \] \[ \frac{3hc}{\lambda} = \phi + K_{\text{max}}' \quad \text{(Equation 3)} \] 6. **Solve for the New Maximum Kinetic Energy ($K_{\text{max}}'$):** Substitute the expression for $\phi$ from Equation 2 into Equation 3: \[ \frac{3hc}{\lambda} = \left( \frac{hc}{\lambda} - K_{\text{max}} \right) + K_{\text{max}}' \] Rearrange the equation to solve for $K_{\text{max}}'$: \[ K_{\text{max}}' = \frac{3hc}{\lambda} - \frac{hc}{\lambda} + K_{\text{max}} \] \[ K_{\text{max}}' = \frac{2hc}{\lambda} + K_{\text{max}} \] Therefore, the new maximum kinetic energy of the photoelectrons is $K_{\text{max}} + \frac{2hc}{\lambda}$. **Comparison with Options:** * (A) $K_{\text{max}} + \frac{2hc}{\lambda}$: This matches our derived result. * (B) $K_{\text{max}} + \frac{hc}{3\lambda}$: This would imply a smaller change in kinetic energy than expected from a shorter wavelength. * (C) $3K_{\text{max}}$: This option incorrectly assumes that kinetic energy scales directly with the inverse of wavelength, ignoring the work function. * (D) $3K_{\text{max}} + \frac{hc}{\lambda}$: This option involves an incorrect combination of terms. The final answer is $\boxed{\text{(A)}}$