bei48482_FM

Several alpha-radioactive nuclides whose atomic numbers are less than 82 are found

in nature, though they are not very abundant.

The intermediate members of each decay series have much shorter half-lives than

their parent nuclide. As a result, if we start with a sample of NAnuclei of a parent

nuclide A, after a period of time an equilibrium situation will come about in which

each successive daughter B,C,... decays at the same rate as it is formed. Thus the

activities RA,RB,RC,... are all equal at equilibrium, and since R Nwe have

NAANBBNCC... (12.9)

Each number of atoms NA,NB,NC,... decreases exponentially with the decay con-

stant Aof the parent nuclide, but Eq. (12.9) remains valid at any time. Equation (12.9)

can be used to establish the decay constant (or half-life) of any member of the series

if the decay constant of another member and their relative proportions in a sample are

known.

Example 12.6

The atomic ratio between the uranium isotopes^238 U and^234 U in a mineral sample is found to

be 1.8 104. The half-life of^234 U is T 1  2 (234) 2.5  105 y. Find the half-life of^238 U.

Solution

Since T 1  2 0.693, from Eq. (12.9) we have

T 1  2 (238) T 1  2 (234)

(1.8 104 )(2.5 105 y)4.5 109 y

This method is convenient for finding the half-lives of very long-lived and very short-lived

radionuclides that are in equilibrium with other radionuclides whose half-lives are easier to measure.

12.4 ALPHA DECAY

Impossible in classical physics, it nevertheless occurs

Because the attractive forces between nucleons are of short range, the total binding

energy in a nucleus is approximately proportional to its mass number A, the number

of nucleons it contains. The repulsive electric forces between protons, however, are of

unlimited range, and the total disruptive energy in a nucleus is approximately

proportional to Z^2 [Eq. (11.12)]. Nuclei which contain 210 or more nucleons are so

large that the short-range nuclear forces that hold them together are barely able to

counterbalance the mutual repulsion of their protons. Alpha decay occurs in such nuclei

as a means of increasing their stability by reducing their size.

Why are alpha particles emitted rather than, say, individual protons or^32 He nuclei?

The answer follows from the high binding energy of the alpha particle. To escape from

a nucleus, a particle must have kinetic energy, and only the alpha-particle mass is

sufficiently smaller than that of its constituent nucleons for such energy to be available.

N(238)



N(234)

Radioactive

equilibrium

432 Chapter Twelve

bei48482_ch12.qxd 1/23/02 12:07 AM Page 432 RKAUL-9 RKAUL-9:Desktop Folder: