Charge Is Conserved — Question 5

Section 21-5 · Physics pages 3460–3461

Question 5 Annihilation changes particle counts while preserving total electric charge An isolated model collection initially contains eight electrons and three positrons. Each electron has charge −e, each positron +e. Three electron–positron pairs annihilate, with each pair producing two neutral photons. Assume no other charged particles and no particles crossing the system boundary. e− e+ γ γ Schematic; labels, not drawing lengths, specify the data. (a) Find initial and final total charge, and the remaining charged-particle and photon counts. (b) For a general initial pair of counts N−, N+ and a annihilations, prove the total charge is independent of a. State the allowed range of a.
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Question 5 - Solution Annihilation changes particle counts while preserving total electric charge e− e+ γ γ (a) Method and calculation Qi = (−8 + 3)e = −5e. N−,f = 8 −3 = 5, N+,f = 3 −3 = 0, Nγ,f = 2(3) = 6. Qf = −5e + 6(0) = −5e. (b) Interpretation and further calculation Qf = −(N−−a)e + (N+ −a)e = (N+ −N−)e = Qi. 0 ≤a ≤min(N−, N+), a ∈Z. Particle number is not the conserved quantity here. Equal positive and negative charges disappear together, leaving their signed sum unchanged. Physical check The neutral photons carry energy and momentum even though they contribute zero electric charge.

Question 5

Annihilation changes particle counts while preserving total electric charge

An isolated model collection initially contains eight electrons and three positrons. Each electron has charge −e-e, each positron +e+e. Three electron–positron pairs annihilate, with each pair producing two neutral photons. Assume no other charged particles and no particles crossing the system boundary.

Diagram shown in the original SVG worksheet.

  1. Find initial and final total charge, and the remaining charged-particle and photon counts.

  2. For a general initial pair of counts N−,N+N_-,N_+ and aa annihilations, prove the total charge is independent of aa. State the allowed range of aa.

Question 5 - Solution

Annihilation changes particle counts while preserving total electric charge

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Qi=(−8+3)e=−5e.Q_i=(-8+3)e=-5e.N−,f=8−3=5,N+,f=3−3=0,Nγ,f=2(3)=6.N_{-,f}=8-3=5,\quad N_{+,f}=3-3=0,\quad N_{\gamma,f}=2(3)=6.Qf=−5e+6(0)=−5e.Q_f=-5e+6(0)=-5e.

(b) Interpretation and further calculation

Qf=−(N−−a)e+(N+−a)e=(N+−N−)e=Qi.Q_f=-(N_--a)e+(N_+-a)e=(N_+-N_-)e=Q_i.0≤a≤min⁡(N−,N+),a∈ℤ.0\le a\le\min(N_-,N_+),\qquad a\in\mathbb Z.Particle number is not the conserved quantity here. Equal positive and negative charges disappear together, leaving their signed sum unchanged.

Physical check

The neutral photons carry energy and momentum even though they contribute zero electric charge.