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Chapter 12

Q. 12.5

Show that the nuclide { }^{8} Be has a positive binding energy but is unstable with respect to decay into two alpha particles.

Strategy We use Equation (12.10) to find the binding energy of { }^{8} Be and Equation (12.16) to find the binding energy of { }^{8} Be with respect to decay to two alpha particles. The nuclide will be unstable if the binding energy B is negative.

B\left({ }_{Z}^{A} X\right)=\left[N m_{n}+Z M\left({ }^{1} H \right)-M\left({ }_{Z}^{A} X\right)\right] c^{2} (12.10)

B=\left[M(R)+M(S)-M\left({ }_{Z}^{A} X\right)\right] c^{2} (12.16)

Step-by-Step

Verified Solution

The binding energy of { }^{8} Be is

 

B\left({ }^{8} Be \right)=\left[4 m_{n}+4 M\left({ }^{1} H \right)-M\left({ }^{8} Be \right)\right] c^{2}

 

We look up the atomic masses of { }^{1} H \text { and }{ }^{8} Be in Appendix 8 and calculate the binding energy to be

 

\begin{aligned}B\left({ }^{8} Be \right) &=[4(1.008665 u )+4(1.007825 u )-8.005305 u ] \\& \times c^{2}\left(\frac{931.5 MeV }{c^{2} \cdot u }\right)=56.5 MeV\end{aligned}

 

So { }^{8} Be has a positive binding energy. Now we calculate the binding energy of the decay of { }^{8} Be into two α particles, { }^{8} Be \rightarrow 2 \alpha, by using Equation (12.16):

 

\begin{aligned}B\left({ }^{8} Be \rightarrow 2 \alpha\right) &=\left[2 M\left({ }^{4} He \right)-M\left({ }^{8} Be \right)\right] c^{2} \\&=[2(4.002603 u )-8.005305 u ] \\& \times c^{2}\left(\frac{931.5 MeV }{c^{2} \cdot u }\right)=-0.093 MeV\end{aligned}

 

Because the latter B is negative, { }^{8} Be is unstable against decay to two alpha particles. From the standpoint of energy, there is no reason why a { }^{8} Be nucleus will not decay into two alpha particles. Sometimes a nuclide may be stable even if another combination of A nucleons has a lower mass, because some conservation law, such as spin angular momentum, prevents the radioactive decay. But in this case we find experimentally that { }^{8} Be does spontaneously decay into two alpha particles. The instability of { }^{8} Be is responsible for the fact that stars consist mostly of hydrogen and helium. Because of the instability of { }^{8} Be, it is difficult for helium nuclei to join together to make heavier nuclei.