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 )

The photoelectric effect is described by the equation: $$K_{\text{max}} = h\nu - \phi$$ where $$K_{\text{max}}$$ is the maximum kinetic energy of ejected electrons, $$h$$ is Planck’s constant, $$\nu$$ is the frequency of incident light, and $$\phi$$ is the work function of the metal.

Since frequency $$\nu$$ is related to wavelength $$\lambda$$ by $$\nu = \frac{c}{\lambda}$$ (where $$c$$ is the speed of light), the equation can be rewritten as: $$K_{\text{max}} = \frac{hc}{\lambda} - \phi$$.

In the first case, when the incident light has wavelength $$\lambda$$, the work function is $$\phi = 1 \text{ eV}$$ and the observed maximum kinetic energy is $$K_1 = 2 \text{ eV}$$. Substituting these values into the equation gives $$2 = \frac{hc}{\lambda} - 1$$, which rearranges to $$2 + 1 = \frac{hc}{\lambda}$$ and hence $$3 = \frac{hc}{\lambda}$$. Thus, $$\frac{hc}{\lambda} = 3 \text{ eV}$$ (1).

Next, when the wavelength of the incident light is halved to $$\frac{\lambda}{2}$$ while the work function remains $$\phi = 1 \text{ eV}$$, let the new maximum kinetic energy be $$K_2$$. The photoelectric equation then yields $$K_2 = \frac{hc}{\frac{\lambda}{2}} - \phi$$, which simplifies to $$K_2 = \frac{2hc}{\lambda} - \phi$$. Substituting $$\frac{hc}{\lambda} = 3 \text{ eV}$$ from (1) and $$\phi = 1 \text{ eV}$$ leads to $$K_2 = 2 \times 3 - 1$$, giving $$K_2 = 6 - 1$$ and therefore $$K_2 = 5 \text{ eV}$$.

Therefore, the maximum kinetic energy of ejected electrons when the surface is illuminated by light of wavelength $$\frac{\lambda}{2}$$ is 5 eV.

The options are: A. 3 eV B. 2 eV C. 6 eV D. 5 eV. The correct answer is D. 5 eV.