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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 1.9×1015Hz?

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Answer #1

The threshold frequency of of cesium = 9.39 x 10^14 Hz

K.E = hf - threshold energy

K.E = hf -hfo (where fo = threshold frequency)

K.E = h(f-fo) = 6.626 E-34 Js (1.9×10^15Hz - 9.39 x 10^14 Hz)

= 6.37 x 10^ -19 J

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