QJEEM25Ph950e: JEE Main 2025 Physics - Class 12 JEEM25Ph950e Solution: A light source of wavelength illuminates a metal surface and…

Let's break down the problem step by step.

The photoelectric equation is given by:

$$K_{\text{max}}=\frac{hc}{\lambda }-\varphi$$

where:

$$K_{\text{max}}$$ is the maximum kinetic energy of the electrons.

$$\frac{hc}{\lambda }$$ is the energy of the incident photon.

$$\varphi$$ is the work function of the metal.

For the initial light source of wavelength $$\lambda$$, we know:

$$2\text{ eV}=\frac{hc}{\lambda }-1\text{ eV}$$

Solving for $$\frac{hc}{\lambda }$$ gives:

$$\frac{hc}{\lambda }=2\text{ eV}+1\text{ eV}=3\text{ eV}$$

Now, if the wavelength is halved to $$\frac{\lambda }{2}$$, the photon energy becomes:

$$\frac{hc}{\lambda /2}=\frac{2hc}{\lambda }=2\times 3\text{ eV}=6\text{ eV}$$

The maximum kinetic energy for the new wavelength is then:

$$K_{\text{max}}^{\prime }=6\text{ eV}-1\text{ eV}=5\text{ eV}$$

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.

The correct answer is Option A.