When UV light of wavelength $$300 \text{ nm}$$ is incident on the metal surface having work function $$2.13 \text{ eV}$$, electron emission takes place. The stopping potential is: (Given $$hc = 1240 \text{ eVnm}$$)

We need to find the stopping potential when UV light of wavelength 300 nm falls on a metal with work function 2.13 eV.

Key Formula: The photoelectric equation (Einstein’s equation) is:

$$ E = W + eV_0 $$

Here $$E = \frac{hc}{\lambda}$$ is the energy of the incident photon, $$W$$ is the work function of the metal, and $$V_0$$ is the stopping potential. Rearranging this equation yields:

$$ eV_0 = \frac{hc}{\lambda} - W $$

The energy of the incident photon is calculated as:

$$ E = \frac{hc}{\lambda} = \frac{1240 \, \text{eV}\cdot\text{nm}}{300 \, \text{nm}} = 4.133 \, \text{eV} $$

Substituting into the photoelectric equation to determine the stopping potential gives:

$$ eV_0 = E - W = 4.133 - 2.13 = 2.003 \, \text{eV} $$

Since $$eV_0$$ is expressed in electronvolts, the stopping potential $$V_0$$ is approximately 2 V.

The correct answer is Option C: 2 V.