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29 December, 08:00

When a certain metal is illuminated with light of frequency 3.0*1015 hz, a stopping potential of 7.27 v is required to stop the most energetic electrons. what is the work function of this metal?

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  1. 29 December, 08:25
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    Hf = Ф + Kmax

    Where,

    h = 4.14*10^-15 eV. s

    f = 3.0*10^15 Hz

    Kmax = 7.27 eV

    Ф = ?

    Therefore,

    Ф = hf-Kmax = 4.14*10^-5*3.0*10^15 - 7.27 = 5.15 eV
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