Given,
ϕ=2.4 eV E=3.3 eV And, E−ϕ=KEmax=12m×v2max
vmax=√0.9×1.6×10−19×29×10−31
vmax=√288×10−109
vmax=√32×10−10
vmax=5.65×10−5
Hence option A is correct.
A photosensitive metallic surface has work function hv0. If photons of energy 2hv0 fall on this surface the electrons come out with a maximum velocity of 4×106 m/s. When the photon energy is increases to 5hv0 then maximum velocity of photo electron will be
When radiation is incident on a photoelectron emitter, the stopping potential is found to be 9 volts. If e/m for the electron is 1.8×1011 Ckg−1 the maximum velocity of the ejected electrons is