Under certain circumstances, a nucleus can decay by emitting a pa… - Vidyadip

Question

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

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Answer

Take a $^{14}_{6}\text{C}$ emission nuclear reaction:
$^{223}_{88}\text{Ra}\rightarrow^{209}_{82}\text{Pb}+^{14}_{6}\text{C}$
We know that:
Mass of $^{223}_{88}\text{Ra},\ \text{m}_1=223.01850\text{ u}$
Mass of $^{209}_{82}\text{Pb},\ \text{m}_2=208.98107\text{ u}$
Mass of $^{14}_{6}\text{C },\text{m}_3=14.00324\text{ u}$
Hence, the Q-value of the reaction is given as:
$\text{Q}=(\text{m}_1-\text{m}_2-\text{m}_3)\text{c}^2$
= (223.01850 - 208.98107 - 14.00324)c²
= (0.03419 c²) u
But 1 u = 931.5 MeV/c²
$\therefore\ \text{Q}=0.03419\times931.5$
= 31.848 MeV
Hence, the Q-value of the nuclear reaction is 31.848 MeV. Since the value is positive, the reaction is energetically allowed.
Now take a $^{4}_{2}\text{He}$ emission nuclear reaction:
$^{223}_{88}\text{Ra}\rightarrow^{219}_{86}\text{Rn}+^{4}_{2}\text{He}$
We know that:
Mass of $^{223}_{88}\text{Ra},\ \text{m}_1=223.01850$
Mass of $^{219}_{82}\text{Rn},\ \text{m}_2=219.00948$
Mass of $^{4}_{2}\text{He },\text{m}_3=4.00260$
Q-value of this nuclear reaction is given as:
$\text{Q}=(\text{m}_1-\text{m}_2-\text{m}_3)\text{c}^2$
= (223.01850 - 219.00948 - 4.00260)C²
= (0.00642 c²) u
= 0.00642 × 931.5 = 5.98 MeV
Hence, the Q value of the second nuclear reaction is 5.98 MeV. Since the value is positive, the reaction is energetically allowed.

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