1. Use Einstein’s photoelectric equation

   $K_{\max} = h\nu - \phi = \frac{hc}{\lambda} - \phi$

   where:

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

2. Given first situation

   For wavelength $\lambda$:

   $K_{\max} = 2\text{ eV}$ $\phi = 1\text{ eV}$

   So photon energy is:

   $\frac{hc}{\lambda} = K_{\max} + \phi = 2 + 1 = 3\text{ eV}$

3. Second situation: wavelength becomes $\frac{\lambda}{2}$

   Photon energy is inversely proportional to wavelength, so halving the wavelength doubles the photon energy:

   $\frac{hc}{\lambda/2} = 2\cdot \frac{hc}{\lambda} = 2\times 3 = 6\text{ eV}$

4. Find new maximum kinetic energy

   $K'_{\max} = 6 - 1 = 5\text{ eV}$

5. Check options

   • A: $5\text{ eV}$ ✅
   • B: $3\text{ eV}$
   • C: $2\text{ eV}$
   • D: $6\text{ eV}$

Therefore, the correct answer is A.