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5 October, 09:03

What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays of frequency 1.6*1015Hz?

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  1. 5 October, 11:18
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    To answer this question, we will use the following formula:

    hf = phi + Emax

    where

    h = plancks constant = 6.626x10^-34 J s

    f = frenquency of incident light = 1.6*10^15 Hz

    phi = work function of the cesium = 2.14 eV = 3.428x10^-19 J

    Emax = maximum kinetic energy of the emitted electrons.

    Substitute with the givens in the above equation to calculate Emax as follows:

    6.626x10^-34 x 1.6*10^15 = 3.428x10^-19 + Emax

    Emax = 7.1736 x 10^-19 J
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