bei48482_FM

The flux of the beam is the number of incident particles per unit area per unit

time, so AN 0  tis their number per unit time. Because Axis the volume of the

sample, the total number of atoms it contains is nnAx. The reaction rate is there-

fore just

Reaction rate ( A)(nx) n (12.22)

Example 12.10

Natural gold consists entirely of the isotope^19779 Au whose cross section for thermal neutron

capture is 99 b. When^19779 Au absorbs a neutron, the product is^19879 Au which is beta-radioactive

with a half-life of 2.69 d. How long should a 10.0-mg gold foil be exposed to a flux of 2.00 

1016 neutrons/m^2  s in order for the sample to have an activity of 200 Ci? Assume that the

irradiation period is much shorter than the half-life of^19879 Au so the decays that occur during

the irradiation can be neglected.

Solution

The decay constant of^19879 Au is

2.98 106 s^1

The required activity of RN 200 Ci2.00 10 ^4 Ci means that the number of^19879 Au

atoms must be

N2.48 1012 atoms

The number of atoms in 10.0 mg 1.00  10 ^5 kg of^19779 Au is

n3.06 1019 atoms

From Eq. (12.22) we find that

t

409 s6 min 49 s

As we assumed, t T 1  2.

12.8 NUCLEAR REACTIONS

In many cases, a compound nucleus is formed first

When two nuclei come close together, a nuclear reaction can occur that results in new

nuclei being formed. Nuclei are positively charged and the repulsion between them

keeps them beyond the range where they can interact unless they are moving very fast

to begin with. In the sun and other stars, whose internal temperatures range up to

2.48^1012 atoms

(2.00 1016 neutronsm^2 s)(3.06 1024 atoms)(99 10 ^28 m^2 )

N



n

1.00 10 ^5 kg



(197 u/atom)(1.66 10 ^27 kgu)

(2.00 10 ^4 Ci)(3.70 1010 s^1 Ci)



2.98 106 s^1

R





0.693



(2.69 d)(86,400 sd)

N



t

446 Chapter Twelve

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