An Interlude — Question 3

Section 44-2 · Physics pages 7316–7317

Question 3 Antiparticle quantum numbers reverse while mass and spin remain equal A hypothetical particle X has charge +2e, baryon number B=1, and spin 1/2. Its antiparticle has the opposite additive charges and the same mass and spin. Consider a system of three X particles and two antiparticles. 3X 2 ¯X X + neutral-charge products Schematic; labels, not drawing lengths, specify the data. (a) Find the total electric charge and baryon number. (b) If all available particle-antiparticle pairs annihilate into states with zero net additive charges, identify the unmatched inventory and verify conservation.
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Question 3 - Solution Antiparticle quantum numbers reverse while mass and spin remain equal 3X 2 ¯X X + neutral-charge products (a) Method and calculation Q ¯ X = −2e, B ¯ X = −1, s ¯ X = 1/2 . Qinitial = 3(2e) + 2(−2e) = 2e, Binitial = 3 −2 = 1 . (b) Interpretation and further calculation 3X + 2 ¯X = X + 2(X ¯X) ⇒one unmatched X . Qfinal = 2e, Bfinal = 1 including all annihilation products. Physical check The bookkeeping does not specify which annihilation channels are dynamically possible. A neutral particle need not be its own antiparticle if other charges distinguish the pair.

Question 3

Antiparticle quantum numbers reverse while mass and spin remain equal

A hypothetical particle X has charge +2e, baryon number B=1, and spin 1/2. Its antiparticle has the opposite additive charges and the same mass and spin. Consider a system of three X particles and two antiparticles.

Diagram shown in the original SVG worksheet.

  1. Find the total electric charge and baryon number.

  2. If all available particle-antiparticle pairs annihilate into states with zero net additive charges, identify the unmatched inventory and verify conservation.

Question 3 - Solution

Antiparticle quantum numbers reverse while mass and spin remain equal

Diagram shown in the original SVG worksheet.

(a) Method and calculation

QX‾=−2e,BX‾=−1,sX‾=1/2.\boxed{Q_{\bar X}=-2e,\quad B_{\bar X}=-1,\quad s_{\bar X}=1/2}.Qinitial=3(2e)+2(−2e)=2e,Binitial=3−2=1.\boxed{Q_{\mathrm{initial}}=3(2e)+2(-2e)=2e,\quad B_{\mathrm{initial}}=3-2=1}.

(b) Interpretation and further calculation

3X+2X‾=X+2(XX‾)⇒one unmatched X.3X+2\bar X=X+2(X\bar X)\Rightarrow\boxed{\text{one unmatched }X}.Qfinal=2e,Bfinal=1including all annihilation products.\boxed{Q_{\mathrm{final}}=2e,\quad B_{\mathrm{final}}=1}\quad\text{including all annihilation products}.

Physical check

The bookkeeping does not specify which annihilation channels are dynamically possible. A neutral particle need not be its own antiparticle if other charges distinguish the pair.