Step 1: Concept
According to Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons is given by $K_{\max} = E - \phi$, where $E$ is the photon energy and $\phi$ is the work function. The stopping potential $V_{0}$ is related to $K_{\max}$ by $K_{\max} = eV_{0}$.
Step 2: Meaning
The energy of the incident photon in electron-volts can be calculated using the short formula $E = \frac{12400}{\lambda\ (\text{in \AA})}$.
Step 3: Analysis
Given $\lambda = 2000\text{ \AA}$ and $\phi = 5.01\text{ eV}$:
$$E = \frac{12400}{2000} = 6.2\text{ eV}$$
Now, finding the maximum kinetic energy:
$$K_{\max} = 6.2\text{ eV} - 5.01\text{ eV} = 1.19\text{ eV} \approx 1.2\text{ eV}$$
Since $K_{\max} = eV_{0}$, the stopping potential is $V_{0} = 1.2\text{ V}$.
Step 4: Conclusion
The stopping potential difference required is 1.2 V.
Final Answer: (A)