Problems#1
A neutral pion at rest decays into two photons. Find the energy, frequency, and wavelength of each photon. In which part of the electromagnetic spectrum does each photon lie?
Answer:
The mass of the pion is 270me, so the rest energy of the pion is
E = mc2 = 270mec2 = 270(0.511MeV) = 138 MeV
Each photon has half this energy, or 69 MeV.
the frequanecy of the photon is
f = E/h = 69 MeV/(6.62 x 10-34 Js)
f = (69 x 106 x 1.602 x 10-19 J)/(6.62 x 10-34 Js)
f = 1.7 x 1022 Hz
and the wavelength of the photon is
c = λf
λ = (2.998 x 108 m/s)/(1.7 x 1022 Hz) = 1.8 x 10-14 m
Problems#2
Two equal-energy photons collide head-on and annihilate each other, producing a pair. The muon mass is given in terms of the electron mass in Section 44.1. (a) Calculate the maximum wavelength of the photons for this to occur. If the photons have this wavelength, describe the motion of the and
immediately after they are produced. (b) If the wavelength of each photon is half the value calculated in part (a), what is the speed of each muon after they have moved apart? Use correct relativistic
expressions for momentum and energy.
Answer:
γ + γ → µ+ + µ-
Each photon must have energy equal to the rest mass energy of a µ+ or µ- is Eµ = 105.7 MeV, then
(a) the maximum wavelength of the photons for this to occur is
Eµ = hc/λ
105.7 x 106 x 1.602 x 10-19 J = (6.62 x 10-34 Js)(3.00 x 108 m/s)/λ
λ = 1.17 x 10-14 m = 0.0117 pm
(b) If the wavelength of each photon is half the value λ/2 so each photon has energy
Eµ = 2 x 105.7 MeV = 211.4 MeV
γmc2 = 211.4 MeV
γ = 211.4 MeV/mc2 = 211.4 MeV/105.7MeV
[1 – (v/c)2]-1/2 = 2
1 – (v/c)2 = ¼
v = ½c√3
Problems#3
A positive pion at rest decays into a positive muon and a neutrino. (a) Approximately how much energy is released in the decay? (Assume the neutrino has zero rest mass. Use the muon and
pion masses given in terms of the electron mass in Section 44.1.) (b) Why can’t a positive muon decay into a positive pion?
Answer:
The mass of the pion is mπ+ = 270me and the mass of the muon is m mμ+ = 207me. The rest
energy of an electron is 0.511 MeV.
(a) ∆m = mπ+ - mμ+ = 270me – 207me = 63me, then
E = γmc2 = 63mec2
E = 63(0.511MeV) = 32 MeV
(b) A positive muon has less mass than a positive pion, so if the decay from muon to pion
was to happen, you could always find a frame where energy was not conserved. This cannot occur.
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