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ChemistryChemistry
10 January, 19:33

What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays of frequency 1.50*1015Hz1.50*1015Hz? Express your answer in joules to three significant figures.

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  1. R
    10 January, 20:04
    0
    Kinetic Energy = 6.51x10^-19 J (3 s. f)

    Explanation:

    frequency = 1.50*1015 s^-1

    Kinetic Energy = ?

    Kinetic Energy = ħf - ф

    h = plank's constant = 6.626x10^-34 Js

    hf = 6.626x10^-34 * 1.50*10^15 = 9.939x10^-19 J

    Work function, ф = 2.1 eV = 3.3642x10^-19 J (upon conversion to joules

    Kinetic energy = 9.939x10^-19 J - 3.3642x10^-19 J

    Kinetic Energy = 6.511x10^-19 J

    Kinetic Energy = 6.51x10^-19 J (3 s. f)
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