Question

a) A photoelectric surface has a work function of 2.75 eV. What is the minimum frequency of light that will cause photoelectron emission from this surface ? answer in the format of a.bc x 10^de Hz

b) A photoelectric cell is illuminated with white light (wavelengths from 400 nm to 700 nm). What is the maximum kinetic energy (in eV) of the electrons emitted by this surface if its work function is 2.30 eV ? 4 digit answer

Answer 1

a) We know that the work function

So

Here f is the minimum frequency of the light that will cause philotoelectron emission.

Here h=6.626×10^(-34) Js

so

f=4.4×10^(-19)/6.626×10^(-34)

f=0.664×10^(15)

f=6.64×10^(-14) Hz. Answer

b) We know that the kinetic energy

. ...2

for maximum kinetic energy , Wavelength should be minimum.

here hc=1240eV.nm ,lambda =400nm , phi=2.30 eV

put these values in equation 2

K=1240eV.nm/400nm -2.3eV

​​​​​​K=0.8 eV.