1. Use Einstein’s photoelectric equation

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

   where:

   • $K_{\max}$ = maximum kinetic energy of photoelectrons
   • $h\nu$ = energy of incident photon
   • $\phi$ = work function of the metal

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

2. Calculate the energy of the incident photon

   Given:

   • Wavelength, $\lambda = 400\,\text{nm}$
   • $hc = 1240\,\text{eV·nm}$

   Photon energy: $E = \frac{hc}{\lambda} = \frac{1240}{400}\,\text{eV} = 3.1\,\text{eV}$

3. Find the work function

   Given kinetic energy: $K_{\max} = 1.68\,\text{eV}$

   Therefore, $\phi = 3.1 - 1.68 = 1.42\,\text{eV}$

4. Match with the closest option

   $\phi \approx 1.42\,\text{eV}$

   Closest option is: A: $1.41\,\text{eV}$

5. Option check

   • A: $1.41\,\text{eV}$ ✅ matches
   • B: $1.51\,\text{eV}$ ❌
   • C: $1.68\,\text{eV}$ ❌ this is the kinetic energy, not work function
   • D: $3.09\,\text{eV}$ ❌ this is nearly the photon energy

Hence, the correct answer is A.