Question

Light of 4.5 ev is incident on a cesium surface and stoping potential is 0.25v maximum k.E of emitted electrons is

Answer 1

formula kinetic energy=potential {difference×charge

K.E= 4.5ev×0.25

or ke=4.5×1.6×10^. --19×0.25

so, maximum kinetic energy of electronis1.8×10^_19 joule

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