QJEEM25Ph24a8: JEE Main 2025 Physics - Class 12 JEEM25Ph24a8 Solution: In photoelectric effect an em-wave is incident on a metal…

Let's break down the problem step by step:

In the photoelectric effect, the energy of a photon is given by:

$h\nu = \phi + K_e$

where:

$\phi$ is the work function of the metal,

$K_e$ is the kinetic energy of the ejected electron.

The kinetic energy of the ejected electrons is provided by the stopping potential:

$K_e = eV$

Given that the stopping potential is $V = 2 \text{ V}$ , we have:

$K_e = 2 \text{ eV}$

Now, we can calculate the photon energy:

$h\nu = \phi + K_e = 2.14 \text{ eV} + 2 \text{ eV} = 4.14 \text{ eV}$

The relation between the photon's energy and its wavelength is:

$h\nu = \frac{hc}{\lambda}$

Rearranging for $\lambda$ :

$\lambda = \frac{hc}{h\nu}$

Given that $$hc=1242\text{ eV·nm}$$, substitute the values:

$$\lambda =\frac{1242\text{ eV·nm}}{4.14\text{ eV}}$$

Calculate the wavelength:

$\lambda \approx \frac{1242}{4.14} \approx 300 \text{ nm}$

Thus, the wavelength of the incident electromagnetic wave is approximately 300 nm.

The correct answer is Option C: 300 nm.