The ejection of the photoelectron from the silver metal in the photoelectric effect experiment can be stopped by applying the voltage of 0.35 V when the radiation 256.7 nm is used. Calculate the work function for silver metal.

Solution

We know
Energy of the incident radiation = Work function + KE of photoelectron
$therefore$ Work function = Energy of the incident
radiation - KE of photoelectron ..(1).
Now energy of the incident radiation (E) = hv
$equals space hc over straight lambda$ ..(2)
$Here space space straight h space equals 6.626 space cross times 10 to the power of negative 34 end exponent Js space colon space straight c space equals 3.0 space cross times space 10 to the power of 8 space ms to the power of negative 1 end exponent semicolon space space space space space space space space straight lambda space equals space 256.7 space nm space equals space 256.7 space cross times 10 to the power of negative 9 end exponent space straight m space space space space$
Substituting the values in eq. (2), we have

$therefore$ Energy of incident radiation

$straight E space equals space fraction numerator left parenthesis 6.626 space cross times 10 to the power of negative 34 end exponent space Js right parenthesis space cross times left parenthesis 3 cross times 10 to the power of 8 space ms to the power of negative 1 end exponent right parenthesis over denominator 256.7 space cross times 10 to the power of negative 9 end exponent straight m end fraction space space space equals space 7.74 space cross times 10 to the power of negative 19 end exponent straight J space space space equals space 4.83 space eV space space space space open square brackets because space space 1 space eV space equals space 1.602 space cross times space 10 to the power of negative 19 end exponent straight J close square brackets$
The potential applied gives the kinetic energy to the electron.
Hence, the kinetic energy of the electron = 4.4 eV.
Substituting the values in eq. (1), we have
Work function = 4.83 eV - 0.35 eV
= 4.48 eV