Problem#1
Estimate the energy width (energy uncertainty) of the if its mean lifetime is 7.6 x 10-21 s. What fraction is this of its rest energy?
Answer:
The rest energy of the Ψ is 3097 MeV.
Apply the Heisenberg uncertainty principle in the form
∆E.∆t ≈ ћ/2
Let Δt be the mean lifetime.
∆E(7.6 x 10-21 s) = (1.054 x 10-34 Js)/2
∆E = 6.93 x 10-15 J = 43 keV
Then, fraction is this of its rest energy is
∆E/mΨc2 = 43 keV/3097000 keV = 1.4 x 10-5
Problem#2
The φ meson has mass 1019.4 MeV/c2 and a measured energy width 4.4 MeV/c2 of Using the uncertainty principle, estimate the lifetime of the φ meson.
Answer:
Apply the Heisenberg uncertainty principle in the form ΔE.Δt ≈ ћ/2. Let ΔE be the energy width and let Δt be the lifetime.
∆t = ћ/2∆E
∆t = (1.054 x 10-34 Js)/(2 x 4.4 x 106 eV x 1.602 x 10-19 J/Ev) = 7.5 x 10-23 s
Problem#3
A φ meson (see Problem 2) at rest decays φ → K+ + K–1 via It has strangeness 0. (a) Find the kinetic energy of the K+ meson. (Assume that the two decay products share kinetic energy equally, since their masses are equal.) (b) Suggest a reason the decay φ → K+ + K–1 + π0 has not been observed. (c) Suggest reasons the decays φ → K+ + π –1 and φ → K+ + µ –1 have not been observed.
Answer:
(a) Does the decays φ → K+ + K–1 . The total energy released is the energy equivalent of the mass decrease.
The mass decrease is
∆m = m(φ) – m(K+) – m(K–1)
The energy released then is
E = ∆mc2 = m(φ)c2 – m(K+)c2 – m(K–1)c2
E = 1019.4 MeV – 493.7MeV – 493.7 MeV = 32.0 MeV
The K+ gets half this, 16.0 MeV.
(b) does the decay φ → K+ + K–1 + π0 . The energy equivalent of the K+ + K–1 + π0 is
E = ∆mc2 = m(π0)c2 +m(K+)c2 + m(K–1)c2
E = 135.0 MeV + 493.7MeV + 493.7 MeV = 1122 MeV
This is greater than the energy equivalent of the φ mass. The mass of the decay products would be greater than the mass of the parent particle; the decay is energetically forbidden.
(c) Does the decay φ → K+ + π –1 . The reaction φ → K+ + K –1 is observed. K+ has strangeness 1+
and Κ–1 has strangeness –1, so the total strangeness of the decay products is zero. If strangeness must be conserved we deduce that the φ particle has strangeness zero. π–1 has strangeness 0, so the product K+ + π–1 has strangeness 1. The decay φ → K+ + π–1 violates conservation of strangeness. Does the decay φ → K+ + μ–1 occur? μ–1 has strangeness 0, so this decay would also violate conservation of strangeness.
Problem#4
One proposed proton decay is p+ → e+ + π0 which violates both baryon and lepton number conservation, so the proton lifetime is expected to be very long. Suppose the proton half-life were 1.0 x 1018 y. (a) Calculate the energy deposited per kilogram of body tissue (in rad) due to the decay of the protons in your body in one year. Model your body as consisting entirely of water. Only the two protons in the hydrogen atoms in each H2O molecule would decay in the manner shown; do you see why? Assume that the π0 decays to two rays, that the positron annihilates with an electron, and that all the energy produced in the primary decay and these secondary decays remains in your body.
(b) Calculate the equivalent dose (in rem) assuming an RBE of 1.0 for all the radiation products, and compare with the 0.1 rem due to the natural background and the 5.0-rem guideline for industrial workers. Based on your calculation, can the proton lifetime be as short as 1.0 x 1018 y?
Answer:
Water has a molecular mass of 18.0 x 10-3 kg/mol.
(a) The number of protons in a kilogram is
N = 1kg[(6.022 x 1023 molecules/mol)/(18.0 x 10-3 kg/mol)](2 protons/molecule) = 6.7 x 1025.
Note that only the protons in the hydrogen atoms are considered as possible sources of proton decay. The energy per decay is
mpc2 = 938.3 MeV = 1.503 x 10-10 J, and so the energy deposited in a year, per kilogram, is
6.7 x 1025[ln2/(1.0 x 1018 y)](1 y)(1.503 x 10-10 J) = 7.0 x 10-3 Gy = 0.70 rad.
(b) For an RBE of unity, the equivalent dose is (1)(0.70 rad) 0.70 rem.
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