1. Use Einstein’s photoelectric equation

   $K_{\max} = h\nu - \phi$

   In electron-volt form,

   $K_{\max}(\text{eV}) = \frac{hc}{\lambda} - \phi$

   Given:

   • $\lambda = 300\,\text{nm}$
   • $\phi = 2.13\,\text{eV}$
   • $hc = 1240\,\text{eV·nm}$

2. Calculate the energy of the incident photon

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

   $E = 4.133\,\text{eV}$

3. Find the maximum kinetic energy of emitted electrons

   $K_{\max} = 4.133 - 2.13 = 2.003\,\text{eV}$

   $K_{\max} \approx 2.0\,\text{eV}$

4. Relate kinetic energy to stopping potential

   $eV_0 = K_{\max}$

   Since $K_{\max}$ is in eV, the stopping potential in volts is numerically equal:

   $V_0 \approx 2.0\,\text{V}$

5. Check options

   • A: $4\,\text{V}$ ❌
   • B: $2\,\text{V}$ ✅
   • C: $4.1\,\text{V}$ ❌
   • D: $1.5\,\text{V}$ ❌

Therefore, the correct option is B.