QJEEM24Phe2eb: JEE Main 2024 Physics - Class 12 JEEM24Phe2eb Solution: In photoelectric experiment energy of 2.48   eV irradiates …

To find the work function of the photo sensitive material, we can use the photoelectric equation which relates the kinetic energy of the ejected electrons to the photon energy and the work function ( $\phi$ ) of the material:

$KE_{\text{max}} = h\nu - \phi$

Where $KE_{\text{max}}$ is the maximum kinetic energy of the ejected electrons, $h\nu$ is the energy of the incoming photon, and $\phi$ is the work function of the material.

However, the kinetic energy of the ejected electrons can also be related to the stopping potential ( $V_s$ ) by the equation:

$KE_{\text{max}} = e \cdot V_s$

Substituting this into the first equation gives:

$e \cdot V_s = h\nu - \phi$

Here, $e$ is the charge of an electron ( $1.6 \times 10^{-19} \, \text{C}$ ), but since we are dealing with energies in electronvolts (eV), and $1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}$ , we can directly use the values given without converting the units:

$\phi = h\nu - e \cdot V_s$

Where $h\nu = 2.48 \, \text{eV}$ is the energy of the irradiating photons, and $V_s = 0.5 \, \text{V}$ .

Substituting these values in, we get:

$\phi = 2.48 \, \text{eV} - 0.5 \, \text{eV} = 1.98 \, \text{eV}$

Therefore, the work function ( $\phi$ ) of the photo sensitive material is $1.98 \, \text{eV}$ , which corresponds to Option A.