Question

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.

Answer #1

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**

(b)

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

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