Page 44 - Modern physical chemistry

Z.8 The Nuclear Atom

To get n, divide the mass density of cadmium by the mass per mole and multiply by the
number of atoms in a mole:

n = 8.642 g cm.a (6.02 x 10 23 atoms mOI· 1 ) = 4.63 x 10 22 atoms cm.a.
112.41 g mol-I

Substitute this and the given 1110 and x into the formula for (1':
a= -In 1.000 x 10-4 =2.4 x 1O-21cm2
(4.63 x 10 22 atoms cm -3 X 0.083 cm)

or
a = 2400 barns.

2.8 The Nuclear Atom
In the first decade of the twentieth century, the openness of materials to bombarding
particles was discovered. In 1903, P. Lenard studied the penetration of matter by cathode
rays (electrons). He found that they passed through foils and thin sheets with attenua-
tion. In 1909, Ernest Rutherford's students, H. Geiger and E. Marsden, found that alpha
rays readily penetrated thin sheets of material. However, a few of the helium nuclei were
deflected through large angles.
In 1911, Rutherford introduced the nuclear atom to explain these results. According
to this model, an atom consists of a small positive nucleus surrounded by a relatively large
region through which the electrons move. Thus, most of the volume of an atom is open.
In determining how large nuclei are, one may advantageously employ neutrons. Elec-
trostatic effects are thus avoided. And if the projectile energy is not too low, in the region
where excessive absorption occurs, or too high, where the nuclei become transparent,
the neutrons absorbed are those hitting the nucleus.
One does have to allow for the radius of the neutron. If both incident particle and
target nucleus are spherical, every projectile whose center strikes within the area 1rlf of
a nucleus, where
[2.39]

does hit the nucleus. Here RA is the radius of the nucleus while Rb is the radius of the
projectile. See figure 2.6.
A beam of particles with definite momentum p is diffracted by a barrier as a ray with
wavevector k given by formula (2.16). In the setup here, each nucleus diffracts projec-
tiles to about the same extent that it absorbs them, So the total cross section is
a = 21rR2 = 2n{ RA + Rb t . [2.40]

Representative results, obtained with 14- and 25-MeV neutrons, appear in table 2.2.
Parameter Ro is defined by the equation

RA =~A1I3 [2.41]

in which A is the mass number, the integer closest to the atomic mass of the isotope.
The unit of distance is thejermi (fm), defined as 1O- 1 5m. The data show that Ro is nearly