Correct Option: C **Step-by-step Solution:** 1. **Recall Einstein's Photoelectric Equation:** Einstein's photoelectric equation relates the energy of the incident photon, the work function of the metal, and the maximum kinetic energy of the emitted photoelectrons. $E = \phi_0 + K_{max}$ where $E$ is the energy of the incident photon, $\phi_0$ is the work function of the metal, and $K_{max}$ is the maximum kinetic energy of the emitted photoelectrons. 2. **Express Photon Energy in terms of Wavelength:** The energy of a photon can be expressed as $E = \frac{hc}{\lambda}$, where $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength of the incident light. 3. **Apply the equation for the initial condition:** For the initial case, with incident wavelength $\lambda$ and maximum kinetic energy $K_{max}$, we have: $\frac{hc}{\lambda} = \phi_0 + K_{max} \quad \ldots(1)$ 4. **Apply the equation for the new condition:** When the incident wavelength is reduced to $\lambda/2$, let the new maximum kinetic energy be $K'_{max}$. $\frac{hc}{\lambda/2} = \phi_0 + K'_{max}$ This simplifies to: $\frac{2hc}{\lambda} = \phi_0 + K'_{max} \quad \ldots(2)$ 5. **Substitute from the initial condition (1) into the new condition (2):** From equation (1), we know that $\frac{hc}{\lambda} = \phi_0 + K_{max}$. Substitute this into equation (2): $2(\phi_0 + K_{max}) = \phi_0 + K'_{max}$ 6. **Solve for the new maximum kinetic energy, $K'_{max}$:** $2\phi_0 + 2K_{max} = \phi_0 + K'_{max}$ $K'_{max} = 2K_{max} + 2\phi_0 - \phi_0$ $K'_{max} = 2K_{max} + \phi_0$ Therefore, the new maximum kinetic energy of the photoelectrons will be $2K_{max} + \phi_0$. **Evaluation of Options:** * A) $2K_{max}$: This would be true if $\phi_0$ were zero or negligible, which is not stated. * B) $K_{max} + \phi_0$: This is the initial photon energy, not the new kinetic energy. * C) $2K_{max} + \phi_0$: This matches our derived result. * D) $K_{max} + 2\phi_0$: This would imply a different relationship, incorrect for this scenario.