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NCERT Class XII Chapter
Nuclei
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Question : 26 of 31
Marks: +1, -0
Under certain circumstances, a nucleus can decay by emitting a particle more massive than an α-particle. Consider the following decay processes:
 88223Ra\,{}^{223}_{88}\mathrm{Ra} →  82209Pb+ 614C\,{}^{209}_{82}\mathrm{Pb}+\,{}^{14}_{6}\mathrm{C} ,  88223Ra\,{}^{223}_{88}\mathrm{Ra} →  86219Rn+ 24He\,{}^{219}_{86}\mathrm{Rn}+\,{}^{4}_{2}\mathrm{He}
Calculate the Q-values for these decays and determine that both are energetically allowed
Solution:  
Let us calculate Q value for the given decay process.
For first decay process
Q = m ( 88223Ra)−m( 82209Pb)−m( 614C)\left(\,{}^{223}_{88}\mathrm{Ra}\right)-m\left(\,{}^{209}_{82}\mathrm{Pb}\right)-m\left(\,{}^{14}_{6}\mathrm{C}\right)
Q = [223.01850 – 208.98107 – 14.00324] (c2c^{2}) u
= [0.034109] × 931.5 MeV = 31.85 MeV
For the second decay process
Q = m ( 88223Ra)−m( 86219Rn)−m( 24He)\left(\,{}^{223}_{88}\mathrm{Ra}\right)-m\left(\,{}^{219}_{86}\mathrm{Rn}\right)-m\left(\,{}^{4}_{2}\mathrm{He}\right)
Q = [223.01850 – 219.00948 – 4.00260] c2c^{2} u
Q = 0.00642 × 931.5 MeV = 5.98 MeV
Since, Q value is positive in both the cases, hence decay process in both ways are possible.
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