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# 3.3: Deeper Analysis of Nuclear Masses

To analyze the masses even better we use the atomic mass unit (amu), which is 1/12th of the mass of the neutral carbon atom,

$1 \text{ amu} = \frac{1}{12} m_{^{12}\mathrm{C}}.$

This can easily be converted to SI units by some chemistry. One mole of $$^{12}$$C weighs $$0.012\text{ kg}$$, and contains Avogadro’s number particles, thus

$1 \text{ amu} = \frac{0.001}{N_A} \text{ kg} = 1.66054 \times 10^{-27} \text{ kg} = 931.494 \text{MeV}/c^2.$

The quantity of most interest in understanding the mass is the binding energy, defined for a neutral atom as the difference between the mass of a nucleus and the mass of its constituents,

$B(A,Z) = Z M_H c^2 + (A-Z) M_n c^2 - M(A,Z)c^2.$

With this choice a system is bound when $$B>0$$, when the mass of the nucleus is lower than the mass of its constituents. Let us first look at this quantity per nucleon as a function of $$A$$, see Figure $$\PageIndex{1}$$.

This seems to show that to a reasonable degree of approximation the mass is a function of $$A$$ alone, and furthermore, that it approaches a constant. This is called nuclear saturation. This agrees with experiment, which suggests that the radius of a nucleus scales with the 1/3rd power of $$A$$,

$R_{\text{RMS}}\approx 1.1 A^{1/3} \text{ fm}.$

This is consistent with the saturation hypothesis made by Gamov in the 30’s:

saturation hypothesis

As $$A$$ increases the volume per nucleon remains constant.

For a spherical nucleus of radius $$R$$ we get the condition

$\frac{4}{3}\pi R^3 = A V_1,$

or

$R = \left(\frac{V_13}{4\pi}\right)^{1/3} A^{1/3}.$

From which we conclude that

$V_1 = 5.5 \text{ fm}^3$