31.6 Binding energy  (Page 3/6)

What is $\text{BE}/A$ For an alpha particle?

Calculate the binding energy per nucleon of ${}^4\text{He}$ , the $\alpha$ particle.

Strategy

To find $\text{BE}/A$ , we first find BE using the Equation $\text{BE}=\{[\text{Zm}({}^1\text{H})+{\text{Nm}}_n]-m({}^A\text{X})\}{c}^2$ and then divide by $A$ . This is straightforward once we have looked up the appropriate atomic masses in [link] .

Solution

The binding energy for a nucleus is given by the equation

For ${}^4\text{He}$ , we have $Z=N=2$ ; thus,

[link] gives these masses as $m({}^4\text{He})=\text{4.002602 u}$ , $m({}^1\text{H})=\text{1.007825 u}$ , and $m_n=\text{1.008665 u}$ . Thus,

Noting that $\text{1 u}=\text{931}\text{.}\text{5 MeV/}{c}^2$ , we find

Since $A=4$ , we see that $\text{BE}/A$ is this number divided by 4, or

Discussion

This is a large binding energy per nucleon compared with those for other low-mass nuclei, which have $\text{BE}/A\approx \text{3 MeV/nucleon}$ . This indicates that ${}^4\text{He}$ is tightly bound compared with its neighbors on the chart of the nuclides. You can see the spike representing this value of $\text{BE}/A$ for ${}^4\text{He}$ on the graph in [link] . This is why ${}^4\text{He}$ is stable. Since ${}^4\text{He}$ is tightly bound, it has less mass than other $A=4$ nuclei and, therefore, cannot spontaneously decay into them. The large binding energy also helps to explain why some nuclei undergo $\alpha$ decay. Smaller mass in the decay products can mean energy release, and such decays can be spontaneous. Further, it can happen that two protons and two neutrons in a nucleus can randomly find themselves together, experience the exceptionally large nuclear force that binds this combination, and act as a ${}^4\text{He}$ unit within the nucleus, at least for a while. In some cases, the ${}^4\text{He}$ escapes, and $\alpha$ decay has then taken place.