To solve this problem, we need to use the concepts of the photoelectric effect and the de-Broglie wavelength.$\\$Given:$\\$• An electromagnetic wave of wavelength $\lambda$ is incident on a photosensitive surface.$\\$• The work function of the surface is negligible.$\\$• $m$ is the mass of the emitted photoelectron.$\\$We need to find the de-Broglie wavelength $\lambda_d$ of the emitted electron.$\\$Step 1: Calculate the energy of the incident photon.$\\ E = \frac{hc}{\lambda} \\$Since the work function is negligible, this energy is entirely converted into the kinetic energy of the photoelectron.$\\$Step 2: Relate the kinetic energy to the de-Broglie wavelength.$\\$The kinetic energy $K$ of the photoelectron is given by:$\\ K = \frac{1}{2}mv^2 = \frac{hc}{\lambda} \\$The de-Broglie wavelength $\lambda_d$ is given by:$\\ \lambda_d = \frac{h}{mv} \\$Step 3: Express $v$ in terms of $\lambda_d. \\ v = \frac{h}{m\lambda_d} \\$Step 4: Substitute $v$ into the kinetic energy equation.$\\ \frac{1}{2}m \left(\frac{h}{m\lambda_d}\right)^2 = \frac{hc}{\lambda} \\$Simplify the equation:$\\ \frac{h^2}{2m\lambda_d^2} = \frac{hc}{\lambda} \\$Step 5: Solve for $\lambda_d. \\ h^2 = 2m\lambda_d^2 \cdot \frac{hc}{\lambda} \\ \lambda = \left(\frac{2mc}{h}\right) \lambda_d^2 \\$Therefore, the correct expression for the de-Broglie wavelength is given by Option 3:$\\ \lambda = \left(\frac{2mc}{h}\right) \lambda_d^2 \\$This corresponds to Option 3.