What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays of frequency 1.9×1015Hz?
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
What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays...
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