QJEEM24Phf5ae: JEE Main 2024 Physics - Class 12 JEEM24Phf5ae Solution: When UV light of wavelength is incident on the metal surface…

To find the stopping potential ( $V_s$ ) when UV light of wavelength $300$ nm is incident on a metal surface with a work function of $2.13$ eV, we can use the photoelectric equation which relates the energy of the incident photons, the work function of the metal, and the kinetic energy of the emitted electrons.

The energy (E) of the photons can be calculated using the equation:

$E = \frac{hc}{\lambda}$

where $h$ is the Planck constant, $c$ is the speed of light, and $\lambda$ is the wavelength of the incident light. Given that $hc = 1240$ eV nm, we can calculate the energy of the UV light photons directly.

Substituting $hc = 1240$ eV nm and $\lambda = 300$ nm into the equation gives:

$E = \frac{1240 \, \text{eV nm}}{300 \, \text{nm}} = 4.13 \, \text{eV}$

Next, we can calculate the maximum kinetic energy of the emitted electrons using the photoelectric effect equation:

$K_{max} = E - \phi$

where $K_{max}$ is the maximum kinetic energy of the emitted electrons, $E$ is the energy of the incident photons, and $\phi$ is the work function of the metal.

Given $E = 4.13$ eV and the work function $\phi = 2.13$ eV, we have:

$K_{max} = 4.13 \, \text{eV} - 2.13 \, \text{eV} = 2 \, \text{eV}$

The stopping potential ( $V_s$ ) is related to the maximum kinetic energy of the emitted electrons by the equation:

$K_{max} = eV_s$

where $e$ is the elementary charge (the charge of an electron), and $V_s$ is the stopping potential. Since $e = 1$ when using energy in eV and potential in volts, the stopping potential $V_s$ can be directly equated to the kinetic energy in eV:

$V_s = K_{max} = 2 \, \text{V}$

Therefore, the stopping potential is $2$ V, which corresponds to Option B.