Show that ^230 92 U does not decay by emitting a neutron or proton. Strategy The decays in question are (a) ^230 92 U → n + ^229 92 U and (b) ^230 92 U → p + ^229 91 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

Question

Show that [latex]{ }_{92}^{230} U[/latex] does not decay by emitting a neutron or proton.

Strategy The decays in question are (a) [latex]{ }_{92}^{230} U \rightarrow n+{ }_{92}^{229} U[/latex] and (b) [latex]{ }_{92}^{230} U \rightarrow p+{ }_{91}^{229} Pa[/latex]. 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.

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

Accepted Answer

(a) [latex]M\left({ }_{92}^{230} U ight)=230.033927 u ; m_{n}=1.008665 u ; M\left({ }_{92}^{229} U ight)=229.033496 u[/latex].

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

Because [latex]Q<0[/latex], neutron decay is not allowed.

(b) [latex]m\left({ }^{ 1 } H ight)=1.007825 u ; M\left({ }_{91}^{229} Pa ight)=229.032089 u[/latex].

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

Because [latex]Q<0[/latex], 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 [latex]{ }^{230} U[/latex] is stable against nucleon emission.

Question 12.13: Show that ^230 92 U does not decay by emitting a neutron or ...

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