Given, λ=310nm ‌=310×10−9m ‌φ=3.55eV⇒v=x×105m∕s ‌x=? Energy of incident photon, ‌E=‌
hc
λ
=‌
6.626×10−34×3×108
310×10−9
‌E=6.412×10−19J Kinetic energy of photoelectrons, ‌KE=E−φ ‌KE=6.412×10−19−3.55×1.6×10−19 ‌KE=0.732×10−19J Velocity of photoelectrons, v‌=√‌
