2a) When a metal surface is illuminated by light of wavelength
310 nm, the
measured maximum kinetic energy of the emitted electrons is 0.50 eV. Calculate the
metal plate’s work function φ in units of eV.
b) In the rest frame of an ejected electron from the photoelectric experiment in part a),
an incident γ-ray with an energy of 0.25 MeV interacts with the electron. Following
the collision, the γ-ray has a final energy of 0.1 MeV. Calculate the angle between the
γ-ray’s initial direction and its direction after the collision.
From the photoelectric effect,
where work function of the metal
wavelength of the light = 310 nm = 310*10-9 m
E = 6.64*10-34*3*108/ (310*10-9) = 0.0642*10-17 = 6.42*10-19J = 6.42*10-19/(1.6*10-19) eV = 4.012 eV
= 4.0125-0.50 = 3.5125 eV
By applying energy and momentum conservation, the relation between the gamma energy before (Eγ =0.25 MeV) and after (Eγ' = 0.1 MeV) the collision can be written as
me = mass of the electron
c = velocity of light
mec2 = rest mass energy of the electron = 0.511 MeV
= 1- (0.25-0.1)0.511 = 1-0.07665 = 0.92335
Get Answers For Free
Most questions answered within 1 hours.