Deeper analysis of nuclear masses

4.3 Deeper analysis of nuclear masses

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

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

(4.1)

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

1 amu = \frac{0.001}{N_{A}} kg = 1.66054 \times 1 0^{- 27} kg = 931.494 MeV ∕ c^{2} .

(4.2)

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} .

(4.3)

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 Fig. 4.1

mass1

Figure 4.1:

B ∕ A

versus

A

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_{RMS} \approx 1.1 A^{1 ∕ 3} fm .

(4.4)

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

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

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

\frac{4}{3} π R^{3} = A V_{1},

(4.5)

R = {(\frac{V_{1} 3}{4 π})}^{1 ∕ 3} A^{1 ∕ 3} .

(4.6)

From which we conclude that

V_{1} = 5.5 {fm}^{3}

(4.7)

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