Question

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?

Answer 1

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