When UV light of wavelength $300 \mathrm{~nm}$ is incident on the metal surface having work function $2.13 \mathrm{~eV}$, electron emission takes place. The stopping potential is :

(Given hc $=1240 \mathrm{~eV} \mathrm{~nm}$ )

<p>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.</p> <p>The energy (E) of the photons can be calculated using the equation:</p> <p>$E = \frac{hc}{\lambda}$</p> <p>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.</p> <p>Substituting $hc = 1240 $ eV nm and $\lambda = 300 $ nm into the equation gives:</p> <p>$E = \frac{1240 \, \text{eV nm}}{300 \, \text{nm}} = 4.13 \, \text{eV}$</p> <p>Next, we can calculate the maximum kinetic energy of the emitted electrons using the photoelectric effect equation:</p> <p>$K_{max} = E - \phi$</p> <p>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.</p> <p>Given $E = 4.13$ eV and the work function $\phi = 2.13$ eV, we have:</p> <p>$K_{max} = 4.13 \, \text{eV} - 2.13 \, \text{eV} = 2 \, \text{eV}$</p> <p>The stopping potential ($V_s$) is related to the maximum kinetic energy of the emitted electrons by the equation:</p> <p>$K_{max} = eV_s$</p> <p>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:</p> <p>$V_s = K_{max} = 2 \, \text{V}$</p> <p>Therefore, the stopping potential is $2$ V, which corresponds to Option B.</p>