and the amount of lead needed to reduce the exposure of a point source of
radiation by 70%.
Answer
Because the initial beam is reduced by 70%, the remaining beam is 30%.
0.3 = 1 ×e−23.1×x
ln(0.3) =−23.1 ×x
1.20 =23.1 ×x
x=0.052 cm
=0.52 mmThus, 0.52 mm of lead will reduce a beam of 140-keV photons by 70%.
Interaction of Neutrons with Matter
Because neutrons are neutral particles, their interactions in the absorber
differ from those of the charged particles. They interact primarily with the
nucleus of the absorber atom and very little with the orbital electrons. The
neutrons can interact with the atomic nuclei in three ways: elastic scatter-
ing, inelastic scattering, and neutron capture. If the sum of the kinetic ener-
gies of the neutron and the nucleus before collision is equal to the sum of
these quantities after collision, then the interaction is called elastic. If a part
of the initial energy is used for the excitation of the struck nucleus, the col-
lision is termed inelastic. In neutron capture, a neutron is captured by the
absorber nucleus, and a new excited nuclide is formed. Depending on the
energy deposited, an a-particle, a proton, a neutron, or g-rays can be emitted
from the excited nucleus, and a new product nuclide (usually radioactive)
is produced.
Questions
- (a) What is the difference between excitation and ionization?
(b) How are d-rays produced?
(c) How much energy is required on the average to produce an ion
pair in air by charged particles? - Define specific ionization (SI), linear energy transfer (LET), and range
(R).
m===−0 693 0 693
003
23 1^1
..
.
.
HVL
cm68 6. Interaction of Radiation with Matter