1. Use Einstein’s photoelectric equation

   For photoelectric emission, $h\nu = \phi + K_{\max}$ where:

   • $\phi$ = work function of the metal
   • $K_{\max} = eV_s$ = maximum kinetic energy

2. Given data

   • Threshold energy (work function): $\phi = 6.2\,\text{eV}$
   • Stopping potential: $V_s = 5\,\text{V}$

   Hence, $K_{\max} = eV_s = 5\,\text{eV}$

3. Find energy of incident radiation

   $h\nu = \phi + K_{\max} = 6.2 + 5 = 11.2\,\text{eV}$

4. Determine the spectral region

   Use $E = \frac{hc}{\lambda}$ or in convenient form, $\lambda(\text{nm}) = \frac{1240}{E(\text{eV})}$

   So, $\lambda = \frac{1240}{11.2} \approx 110.7\,\text{nm}$

5. Identify region from wavelength

   A wavelength of about $111\,\text{nm}$ lies in the ultra-violet region.

6. Check options

   • A: ultra-violet region ✅ Correct
   • B: infra-red region ❌ Much longer wavelength, lower energy
   • C: visible region ❌ Visible light is about $400$–$700\,\text{nm}$
   • D: $x$-ray region ❌ $x$-rays have much shorter wavelength and much higher energy

Therefore, the correct option is A.