Physics · Ch 8 — Atomic and Nuclear Physics

Size and density of the nucleus

8.4.4

Size and density of the nucleus

The alpha-particle scattering experiment, together with many other independent measurement techniques applied to a wide range of nuclei, shows that nuclei are approximately spherical, and that for nuclei with $Z>10$ the radius $R$ obeys a simple empirical relation to the mass number $A$ :

R=R_0A^{1/3}\qquad (8.19)

where the constant $R_0=1.2$ fermi ( $1\ \text{F}=1\times10^{-15}$ m, the unit named after Enrico Fermi). Example 8.7 applies this directly: for the gold nucleus $^{197}_{79}Au$ ( $A=197$ ), $R=1.2\times(197)^{1/3}\approx6.97$ F.

Constant nuclear density. Because the radius scales as $A^{1/3}$ , the nuclear volume scales directly as $A$ : $V=\tfrac{4}{3}\pi R^3=\tfrac{4}{3}\pi R_0^3A$ . If the small mass difference between protons and neutrons is ignored, the total nuclear mass is approximately $A\,m_p$ , where $m_p$ is the proton mass. The nuclear density is then

$\rho=\frac{\text{mass}}{\text{volume}}=\frac{Am_p}{\tfrac{4}{3}\pi R_0^3A}=\frac{m_p}{\tfrac{4}{3}\pi R_0^3}$ …

Misc Example 8.7Radius of the gold-197 nucleus

Worked out. This worked example applies the empirical radius formula R = R0 A^(1/3), with R0 = 1.2 fermi, to the gold nucleus $^{197}_{79}\text{Au}$ , whose mass number is A = 197. Substituting A = 197 gives R = 1.2 times (197)^(1/3) fermi, and since the cube root of 197 works out to about 5.81, the radius comes out to R = 6.97 fermi, illustrating how directly the mass number alone determines the nuclear size once the emp …

Misc Example 8.8Density of a nucleus with mass number A

Worked out. This worked example derives that nuclear density is independent of mass number: starting from the nuclear volume V = (4/3) pi R^3 = (4/3) pi (R0 A^(1/3))^3 = (4/3) pi R0^3 A, and taking the nuclear mass to be approximately A times the proton mass m_p (ignoring the tiny proton-neutron mass difference), the density rho = mass/volume = (A m_p)/((4/3) pi R0^3 A) has the mass number A cancel out completely, leaving rho = m_p / ((4/3) pi R0^3). Substituting m_p = 1.67e-27 kg and R0 = 1.2e-15 m gives a numerical density of about 2.3e17 kg per cubic metre for every nucleus with Z greater than 10, a value roughly 10^14 times the density of ordinary water, sh …

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