Using Einstein's photoelectric equation:
λhc=Φ+eV0
The energy of the incident photon is:
E=λhc=331×10−96.62×10−34×3×108
E=331×10−919.86×10−26=6×10−19 J
The maximum kinetic energy of the photoelectrons is:
Kmax=eV0=1.6×10−19×0.2=0.32×10−19 J
The work function Φ is:
Φ=E−Kmax
Φ=6×10−19−0.32×10−19=5.68×10−19 J
Comparing with α×10−19 J, we get α=5.68.
Answer: 5.68