Question

( therefore ) Work function ( = ) Energy of the incident radiation - KE of photoelectron
Now energy of the incident radiation ( (mathrm{E})=mathrm{hv} )
[
=frac{h c}{lambda}
]
Here ( h=6.626 times 10^{-34} mathrm{Js}: mathrm{c}=3.0 times 10^{8} mathrm{ms}^{-1} )
[
lambda=256.7 mathrm{nm}=256.7 times 10^{-9} mathrm{m}
]
Substituting the values in eq. (2), we have
( therefore ) Energy of incident radiation
[
begin{aligned}
E &=frac{left(6.626 times 10^{-34} mathrm{Js}right) timesleft(3 times 10^{8} mathrm{ms}^{-1}right)}{256.7 times 10^{-9} mathrm{m}}
&=7.74 times 10^{-19} mathrm{J}
&=4.83 mathrm{eV}left[because 1 mathrm{eV}=1.602 times 10^{-19} mathrm{J}right]
end{aligned}
]
The potential applied gives the kinetic energy to the electron.
Hence, the kinetic energy of the electron ( =4.4 mathrm{eV} ). Substituting the values in eq. (1), we have
[
begin{array}{l}
text { Work function }=4.83 mathrm{eV}-0.35 mathrm{eV}
qquad=4.48 mathrm{eV}
end{array}
]

# 13. The ejection of the photoelectrons 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