A light source of wavelength $\lambda$ illuminates a metal surface and electrons are ejected with maximum kinetic energy of 2 eV . If the same surface is illuminated by a light source of wavelength $\frac{\lambda}{2}$, then the maximum kinetic energy of ejected electrons will be (The work function of metal is 1 eV )

<p>Let's break down the problem step by step.</p> <p><p>The photoelectric equation is given by:</p> <p>$K_{\text{max}} = \frac{hc}{\lambda} - \phi$</p> <p>where:</p></p> <p><p>$K_{\text{max}}$ is the maximum kinetic energy of the electrons.</p></p> <p><p>$\frac{hc}{\lambda}$ is the energy of the incident photon.</p></p> <p><p>$\phi$ is the work function of the metal.</p></p> <p><p>For the initial light source of wavelength $\lambda$, we know:</p> <p>$2 \text{ eV} = \frac{hc}{\lambda} - 1 \text{ eV}$</p> <p>Solving for $\frac{hc}{\lambda}$ gives:</p> <p>$\frac{hc}{\lambda} = 2 \text{ eV} + 1 \text{ eV} = 3 \text{ eV}$</p></p> <p><p>Now, if the wavelength is halved to $\frac{\lambda}{2}$, the photon energy becomes:</p> <p>$$\frac{hc}{\lambda/2} = \frac{2hc}{\lambda} = 2 \times 3 \text{ eV} = 6 \text{ eV}$$</p></p> <p><p>The maximum kinetic energy for the new wavelength is then:</p> <p>$K'_{\text{max}} = 6 \text{ eV} - 1 \text{ eV} = 5 \text{ eV}$</p></p> <p>Therefore, the maximum kinetic energy of the ejected electrons when the surface is illuminated by a light source of wavelength $\frac{\lambda}{2}$ is 5 eV.</p> <p>The correct answer is Option A.</p>