11.5 Substructure of the nucleus  (Page 3/16)

How small and dense is a nucleus?

(a) Find the radius of an iron-56 nucleus. (b) Find its approximate density in $\mathrm{kg}/{\mathrm{m}}^3$ , approximating the mass of ${}^{56}\text{Fe}$ to be 56 u.

Strategy and Concept

(a) Finding the radius of ${}^{56}\text{Fe}$ is a straightforward application of $r=r_0{A}^{1/3},$ given $A=56$ . (b) To find the approximate density, we assume the nucleus is spherical (this one actually is), calculate its volume using the radius found in part (a), and then find its density from $\rho =\mathrm{m/V}$ . Finally, we will need to convert density from units of $\mathrm{u}/{\mathrm{fm}}^3$ to $\mathrm{kg}/{\mathrm{m}}^3$ .

Solution

(a) The radius of a nucleus is given by

Substituting the values for $r_0$ and $A$ yields

(b) Density is defined to be $\rho =\mathrm{m/V}$ , which for a sphere of radius $r$ is

Substituting known values gives

Converting to units of $\mathrm{kg}/{\mathrm{m}}^3$ , we find

Discussion

(a) The radius of this medium-sized nucleus is found to be approximately 4.6 fm, and so its diameter is about 10 fm, or ${10}^{\mathrm{–14}}\text{m}$ . In our discussion of Rutherford’s discovery of the nucleus, we noticed that it is about ${10}^{\mathrm{–15}}\text{m}$ in diameter (which is for lighter nuclei), consistent with this result to an order of magnitude. The nucleus is much smaller in diameter than the typical atom, which has a diameter of the order of ${10}^{\mathrm{–10}}\text{m}$ .

(b) The density found here is so large as to cause disbelief. It is consistent with earlier discussions we have had about the nucleus being very small and containing nearly all of the mass of the atom. Nuclear densities, such as found here, are about $2\times {10}^{14}$ times greater than that of water, which has a density of “only” ${10}^3{\text{kg/m}}^3$ . One cubic meter of nuclear matter, such as found in a neutron star, has the same mass as a cube of water 61 km on a side.