WAEC Physics 2011 Theory — Question 36
Question 36 of 39 from the West African Examinations Council (WAEC) Physics 2011 Theory paper, with the correct answer and a full explanation.
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15(b)(i) In the nuclear equation 23/11 A + 2/1 B -> p/q C + proton, A and B, quickly decays to another nucleus E as indicated in the equations above. Determine the values of p, q, r and s.
Model answer
23/11 A + 2/1 B -> p/q C + 1/1 H (proton). Mass number conservation: 23+2 = p+1, so p = 25-1 = 24. Atomic number conservation: 11+1 = q+1, so q = 12-1 = 11. Therefore p/q C = 24/11 C. p/q C -> r/s E + beta particle (0/-1 B): Since p=24 and q=11, and beta emission increases atomic number by 1 while mass number stays the same: r = 24, s = 11+1 = 12. Therefore r/s E = 24/12 E.
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