Quantum Physics

29.1 Some Properties of Nuclei 941The Size of NucleiThe size and structure of nuclei were first investigated in the scattering experimentsof Rutherford, discussed in Section 28.1. Using the principle of conservationof energy, Rutherford found an expression for how close an alpha particlemoving directly toward the nucleus can come to the nucleus before being turnedaround by Coulomb repulsion.In such a head-on collision, the kinetic energy of the incoming alpha particlemust be converted completely to electrical potential energy when the particlestops at the point of closest approach and turns around (Active Fig. 29.1). If weequate the initial kinetic energy of the alpha particle to the maximum electricalpotential energy of the system (alpha particle plus target nucleus), we have12 mv 2 q 1 q 2 k erwhere d is the distance of closest approach. Solving for d, we getd 4k eZe 2mv 2From this expression, Rutherford found that alpha particles approached to within3.2 10 14 m of a nucleus when the foil was made of gold. Thus, the radius of thegold nucleus must be less than this value. For silver atoms, the distance of closestapproach was 2 10 14 m. From these results, Rutherford concluded that thepositive charge in an atom is concentrated in a small sphere, which he called thenucleus, with radius no greater than about 10 14 m. Because such small lengthsare common in nuclear physics, a convenient unit of length is the femtometer (fm),sometimes called the fermi and defined as1 fm 10 15 mSince the time of Rutherford’s scattering experiments, a multitude of otherexperiments have shown that most nuclei are approximately spherical and havean average radius given byr r 0 A 1/3 k e(2e)(Ze)d[29.1]where A is the total number of nucleons and r 0 is a constant equal to 1.2 10 15 m.Because the volume of a sphere is proportional to the cube of its radius, it followsfrom Equation 29.1 that the volume of a nucleus (assumed to be spherical) isdirectly proportional to A, the total number of nucleons. This relationship thensuggests that all nuclei have nearly the same density. Nucleons combine to form anucleus as though they were tightly packed spheres (Fig. 29.2).Ze2e v = 0++ +++ ++ ++ ++ACTIVE FIGURE 29.1An alpha particle on a head-on collisioncourse with a nucleus of chargeZe. Because of the Coulomb repulsionbetween the like charges, the alphaparticle will stop instantaneously at adistance d from the nucleus, calledthe distance of closest approach.Log into PhysicsNow at www.cp7e.comand go to Active Figure 29.1, whereyou can adjust the atomic number ofthe target nucleus and the kineticenergy of the alpha particle. Thenobserve the approach of the alphaparticle toward the nucleus.Figure 29.2 A nucleus can bevisualized as a cluster of tightly packedspheres, each of which is a nucleon.dEXAMPLE 29.1GoalSizing a Neutron StarApply the concepts of nuclear size.Problem One of the end stages of stellar life is a neutron star, where matter collapses and electrons combine with protonsto form neutrons. Some liken neutron stars to a single gigantic nucleus. (a) Approximately how many nucleons arein a neutron star with a mass of 3.00 10 30 kg? (This is the mass number of the star.) (b) Calculate the radius of thestar, treating it as a giant nucleus. (c) Calculate the density of the star, assuming the mass is distributed uniformly.Strategy The effective mass number of the neutron star can be found by dividing the star mass in kg by the mass ofa neutron. Equation 29.1 then gives an estimate of the radius of the star, which together with the mass determinesthe density.Solution(a) Find the approximate number of nucleons in the star.Divide the star’s mass by the mass of a neutron to find A: 1.79 10 57A 3.00 10 30 kg1.675 10 27 kg