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what is the kinetic energy of the emitted electrons when cesium is exposed to uv rays...

what is the kinetic energy of the emitted electrons when cesium is exposed to uv rays of frequency 1.2×10^15 hz

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Answer #1

frequency = 1.2×10^15 hz

photon energy E = h v

                           = 6.625 x 10^-34 x 1.2 x 10^15

                          = 7.95 x 10^-19 J

work function of Cs = 3.36 x 10^-19 J

K E = E - W

     = 7.95 x 10^-19 - 3.36 x 10^-19

KE = 4.58 x 10^-19 J

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