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3 years ago

10

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?

1 answer:

slega [8]3 years ago

7 0

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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