Is the mass excess of an alpha particle (use a straightedge on Fig. 42-13) greater than or less than the particle’s total binding energy (use the binding energy per nucleon from Fig. 42-7)?

Figure 42-13 and figure 42-7 is given.

The mass excess is an expression of the nuclear binding energy of a particle, relative to the binding energy per nucleon of carbon-12 (which defines the atomic mass unit). If the mass excess is negative, the nucleus has more binding energy than carbon-12 and vice versa.

Let a particle A undergoes alpha decay to convert into B.

That is: ${}_Z^{M}A{\to }_{Z-2}^{M-4}B{+}_2^{4}He(\alpha -particle)$

Again, consider that a binding energy of the particle is given by the mass defect that implies the condition $m_1+m_2<M$ to satisfy for getting the binding energy of the particle, where, M is the mass of A,$m_1$ is the mass of B and $m_2$is the mass of alpha particle.

But the mass excess of an alpha particle is given by the difference between the actual mass and the atomic mass of an alpha particle. That is given by:

$$\begin{gathered} ∆\mathrm{m}=\left(4.0026\mathrm{u}\right)-\left(4\mathrm{u}\right) \\ =0.00260\mathrm{u} \\ =2.4219\mathrm{MeV} \end{gathered}$$

That lay in stable nuclei regions.

Again comparing the values of the mass excess of an alpha particle and the binding energy of any particle from figures 42-13 and 42-7, it is considered that the mass excess is always greater for an alpha particle than the particle’s binding energy per nucleon.