Problem#1
Calculate the threshold kinetic energy for the reaction π– + p → ∑0 + K0 if a π– beam is incident on a stationary proton target. The K0 has a mass of 497.7 MeV/c2.
Answer:
We use
Ea2 = 2Mc2Em + Mc2 + (mc2)2
Where, Ea = (m∑0 + mK0)c2, M = mp, m = mπ- and Em = (mπ-)c2 + K
Then,
Ea2 = 2mpc2Em + (mpc2)2 + (mπ-c2)2
Ea2 = 2mpc2[(mπ-)c2 + K] + (mpc2)2 + (mπ-c2)2
(m∑0c2 + mK0c2)2 = 2mpc2[(mπ-)c2 + K] + (mpc2)2 + (mπ-c2)2
(1193MeV + 497.7 MeV)2 = 2 x 938.3 MeV[139.6MeV + K] + (938.3 MeV)2 + (139.6MeV)2
1043.68 MeV = 139.6MeV + K
K = 904.1 MeV
Problem#2
Calculate the threshold kinetic energy for the reaction p + p → p + p + K+ + K–1 if a proton beam is incident on a stationary proton target.
Answer:
With a stationary target, only part of the initial kinetic energy of the moving proton is available. Momentum conservation tells us that there must be nonzero momentum after the collision, which means that there must also be leftover kinetic energy. Therefore not all of the initial energy is available.
The minimum amount of available energy must be
Ea = 2mpc2 + mK+c2 + mK-c2
Ea = 2(938.3 MeV) + 493.7 MeV + 493.7 MeV = 2864 MeV
Solving the available energy formula for Em gives
Ea2 = 2mc2(Em + mc2)
(2864 MeV)2 = 2(938.3 MeV)(Em + 938.3MeV)
4370.93 MeV = Em + 938.3MeV
Em = 3432.6 MeV
Recalling that Em is the total energy of the proton, including its rest mass energy (RME), we have
K = Em – RME = 3432.6 MeV – 938.3 MeV = 2494 MeV
Therefore the threshold kinetic energy is 2494 MeV
Problem#3
An η0 meson at rest decays into three π mesons. (a) What are the allowed combinations π0, π+ and π–1 of and as decay products? (b) Find the total kinetic energy of the mesons.
Answer:
Charge must be conserved. The energy released equals the decrease in rest energy that occurs in the decay.
(a) The decay products must be neutral, so the only possible combinations are
π0π0π0 or π0 π+ π–1
(b) the total kinetic energy of the mesons is
K = (mη0 - mπ0 - mπ+ - mπ-)c2
K = 547.3 MeV – 135.0 MeV – 139.6 MeV – 139.6 MeV = 133.1 MeV
Problem#4
Each of the following reactions is missing a single particle. Calculate the baryon number, charge, strangeness, and the three lepton numbers (where appropriate) of the missing particle, and from this identify the particle. (a) p + p → p + Λ0 + ?; (b) K–1 + n → Λ0 + ?; (c) p + p̅ → n + ?; (d) v̅µ + p → n + ?.
Answer:
(a) The baryon number is 0, the charge is , +e the strangeness is 1, all lepton numbers are zero, and the particle is K+.
(b) The baryon number is 0, the charge is , -e the strangeness is 0, all lepton numbers are zero and the particle is π–.
(c) The baryon number is –1, the charge is 0, the strangeness is zero, all lepton numbers are 0 and the particle is an antineutron.
(d) The baryon number is 0 the charge is , +e the strangeness is 0, the muonic lepton number is –1, all other lepton numbers are 0 and the particle is μ+.
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