Energy required to remove an electron from aluminium surface is 4.2 eV. If light of wavelength 2000∘A falls on the surface, the velocity of the fastest electron ejected from the surface will be
8.4×105 m/sec
By using E=W0+12mv2max where E=123752000=6.18eV
⇒6.18eV=4.2eV+12mv2max⇒1.98eV=12mv2max
⇒1.98×1.6×10−19=12×9.1×10−31×v2max
⇒vmax=8.4×105 m/s