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Additional Exercises · 13.26

Q.Under certain circumstances, a nucleus can decay by emitting a particle more massive than an α\alpha-particle. Consider the following decay processes:
[!FORMULA] 88223Ra→82209Pb+614C^{223}_{88}\text{Ra} \rightarrow {}^{209}_{82}\text{Pb} + {}^{14}_{6}\text{C}
[!FORMULA] 88223Ra→86219Rn+24He^{223}_{88}\text{Ra} \rightarrow {}^{219}_{86}\text{Rn} + {}^{4}_{2}\text{He}
Calculate the Q-values for these decays and determine that both are energetically allowed.

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Using standard atomic mass values for Ra-223, Pb-209, C-14 and Rn-219 (this exercise's real textbook printing does not restate these masses inline — it relies on the book's own Appendix mass table, unlike most other exercises in this chapter), Q≈31.8Q \approx 31.8 MeV for carbon-14 emission and Q≈5.98Q \approx 5.98 MeV for ordinary alpha emission — both positive, confirming both channels are allowed, though the far larger Q for alpha decay is why it dominates in practice.

Masses used (standard nuclear data, since not given in the exercise's own printed text):

m(223Ra)≈223.018502 u,m(209Pb)≈208.981091 um(^{223}\text{Ra}) \approx 223.018502\ \text{u}, \quad m(^{209}\text{Pb}) \approx 208.981091\ \text{u}

m(14C)≈14.003242 u,m(219Rn)≈219.009480 u,m(4He)=4.002603 u (given)m(^{14}\text{C}) \approx 14.003242\ \text{u}, \quad m(^{219}\text{Rn}) \approx 219.009480\ \text{u}, \quad m(^{4}\text{He}) = 4.002603\ \text{u (given)}

Channel 1: 88223Ra→82209Pb+614C^{223}_{88}\text{Ra} \rightarrow {}^{209}_{82}\text{Pb} + {}^{14}_{6}\text{C}

Q1=[m(223Ra)−m(209Pb)−m(14C)]×931.5Q_1 = \left[m(^{223}\text{Ra}) - m(^{209}\text{Pb}) - m(^{14}\text{C})\right] \times 931.5

=[223.018502−208.981091−14.003242]×931.5= [223.018502 - 208.981091 - 14.003242] \times 931.5

=0.034169×931.5=31.83 MeV= 0.034169 \times 931.5 = 31.83\ \text{MeV}

Channel 2: 88223Ra→86219Rn+24He^{223}_{88}\text{Ra} \rightarrow {}^{219}_{86}\text{Rn} + {}^{4}_{2}\text{He}

Q2=[m(223Ra)−m(219Rn)−m(4He)]×931.5Q_2 = \left[m(^{223}\text{Ra}) - m(^{219}\text{Rn}) - m(^{4}\text{He})\right] \times 931.5

=[223.018502−219.009480−4.002603]×931.5= [223.018502 - 219.009480 - 4.002603] \times 931.5 …

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