Particles and Interactions Problems and Solutions 2

 Problem#1

The discovery of the Ω^– particle helped confirm GellMann’s eightfold way. If an Ω^– decays into a Λ⁰ and a K^–, what is the total kinetic energy of the decay products?

Answer:
If the initial and final rest mass energies were equal, there would be no leftover energy for kinetic energy. Therefore the kinetic energy of the products is the difference between the mass energy of the initial particles and the final particles.

The difference in mass is

∆m = M_Ω – m_Λ – m_K

the energy difference is

E = ∆mc² = (1672 MeV/c² – 1116 MeV/c² – 494 MeV/c²)c² = 62 MeV

There is less rest mass energy after the reaction than before because 62 MeV of the initial

energy was converted to kinetic energy of the products.

Problem#2
In which of the following decays are the three lepton numbers conserved? In each case, explain your reasoning. (a) µ^– → e^– + υ_e + ῡ_µ­, (b) τ^– → e^– + ῡ_e + υ_τ; (c) π⁺ → e⁺ + γ; (d) n → p + e^– + ῡ_e.

Answer:
(a) µ^– → e^– + υ_e + ῡ_µ­

 L_µ: +1 ≠ –1, L_e: 0 ≠ +1 + 1 so lepton numbers are not conserved.

(b) τ^– → e^– + ῡ_e + υ_τ

L_e: 0 = +1 – 1; L_τ: +1 = +1, so lepton numbers are conserved.

(c) π⁺ → e⁺ + γ

Lepton numbers are not conserved since just one lepton is produced from zero original leptons.

(d) n → p + e^– + ῡ_e.

L_e: 0 = +1 – 1; so the lepton numbers are conserved.

Problem#3
Which of the following reactions obey the conservation of baryon number?, (a) p + p → p⁺ + e⁺; (b) p + n → 2e⁺ + e^–; (c) p → n + e^– + ῡ_e.

Answer:
p and n have baryon number +1 and p has baryon number -1. e , e , -υ_e and γ all have baryon number zero. Baryon number is conserved if the total baryon number of the products equals the total baryon number of the reactants.

(a) reactants: B = 1 + 1 = 2. Products: B = 1 + 0 = +1. Not conserved.

(b) reactants: B = 1 + 1 = 2. Products: B = 0 + 0 = 0. Not conserved.

(c) reactants: B = +1. Products: B = 1 + 0 + 0 = +1. Conserved.

(d) reactants: B = 1 – 1. Products: B = 0. Conserved.

Even though a reaction obeys conservation of baryon number it may still not occur

spontaneously, if it is not energetically allowed or if other conservation laws are violated

Problem#4
In which of the following reactions or decays is strangeness conserved? In each case, explain your reasoning. (a) K⁺ → µ⁺ + ν_µ; (b) n + K⁺ → p + π⁰; (c) K⁺ + K^– → π⁰ + π⁰; (d) p + K^– → Λ⁰ + π⁰.

Answer:
Compare the sum of the strangeness quantum numbers for the particles on each side of the decay equation.

(a) K⁺ → µ⁺ + ν_µ;

S_K+ = +1, S_µ+ = 0 and S_vµ = 0, then

S = 1 initially; S = 0 for the products; S is not conserved

(b) n + K⁺ → p + π⁰;

S_n = 0, S_K+ = +1, S_p = 0, and S_π0 = 0, then

S = 1 initially; S = 0 for the products; S is not conserved

(c) K⁺ + K^– → π⁰ + π⁰;

S_K+ = +1, S_K– = –1, and S_π0 = 0, then

S = 1 – 1 = 0, initially; S = 0 for the products; S is conserved

(d) p + K^– → Λ⁰ + π⁰.

S_p = 0, S_K– = –1, S_Λ0 = –1, and S_π0 = 0, then

S = -1 initially; S = -1 for the products; S is conserved

Problem#5
a) Show that the coupling constant for the electromagnetic interaction, e²/4πϵ₀ћc, is dimensionless and has the numerical value (b) Show that in the Bohr model the orbital speed of an electron in the n = 1 orbit is equal to c times the coupling constant e²/4πϵ₀ћc.

Answer:
(a) e²/4πϵ₀ћc = 2π(1,602 x 10^-19C)²(9.0 x 10⁹ Nm²/C²)/[(6.626 x 10^-34 Js)(2.998 x 10⁸ m/s)]
= 7.29660475 x 10^-3

≈ 1/137 to three figures.

(b) Hence Newton’s second law states that

e²/4πϵ₀r_n² = mv_n²/r_n

then, orbital speeds in the Bohr model is

v_n = e²/2nϵ₀h

for, n = 1

v₁ = e²/2ϵ₀h or

v₁ = [e²/4πϵ₀ћc]c

where ћ = h/2π.

Problem#6
Show that the nuclear force coupling constant f²/ћc is dimensionless.

Answer:
f² has units of energy times distance. ћ has units of J.s and c has units of m/s. Then,

f²/ћc = (J.m)/[(J.s)(m.s^-1)] = 1

thus f²/ћc is dimensionless.    

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