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## Q. 12.13

Show that ${ }_{92}^{230} U$ does not decay by emitting a neutron or proton.

Strategy The decays in question are (a) ${ }_{92}^{230} U \rightarrow n+{ }_{92}^{229} U$ and (b) ${ }_{92}^{230} U \rightarrow p+{ }_{91}^{229} Pa$. We determine whether the decays occur by looking up the atomic masses in Appendix 8 and use Equation (12.29b) to see if Q is positive or negative.

$Q=\left[M\left({ }_{Z}^{A} X\right)-M_{D}-M_{y}\right] c^{2}$ (12.29b)

## Verified Solution

(a) $M\left({ }_{92}^{230} U \right)=230.033927 u ; m_{n}=1.008665 u ; M\left({ }_{92}^{229} U \right)=229.033496 u$.

\begin{aligned}Q=&[230.033927 u -229.033496 u -1.008665 u ] c^{2} \\& \times\left(\frac{931.5 MeV }{c^{2} \cdot u }\right)=-7.7 MeV\end{aligned}

Because $Q<0$, neutron decay is not allowed.

(b) $m\left({ }^{ 1 } H \right)=1.007825 u ; M\left({ }_{91}^{229} Pa \right)=229.032089 u$.

\begin{aligned}Q=&[230.033927 u -229.032089 u -1.007825 u ] c^{2} \\& \times\left(\frac{931.5 MeV }{c^{2} \cdot u }\right)=-5.6 MeV\end{aligned}

Because $Q<0$, proton decay is not allowed.

In both cases the decay is not allowed, because the mass of the products is greater than that of the parent. The nucleus ${ }^{230} U$ is stable against nucleon emission.