Physics and Engineering of Radiation Detection

1.3. Radioactivity and Radioactive Decay 15

Since half life represents the time taken by half of the atoms in a sample to

decay, we can simply replaceNbyN 0 /2 in equation 1.3.13 to get

1

2

= e−λdT^1 /^2

eλdT^1 /^2 =2

T 1 / 2 =

ln(2)

λd

=ln(2)τ=0. 693 τ.

Example:

The half life of a radioactive sample is found to be 45 days. How long would

it take for 2 moles of this material to decay into 0.5 mole.

Solution:

SinceT 1 / 2 = 45 days, therefore

λd =

ln(2)

T 1 / 2

=

ln(2)

45

=15. 4 × 10 −^3 day−^1.

Since moleMis proportional to the number of atoms in the material, therefore

equation 1.3.13 can also be written in terms of number of moles as follows.

M=M 0 e−λdt

Rearrangement of this equation gives

t=

1

λd

ln

(

M 0

M

)

.

Hence the time it will take for 1.5 moles of this material to decay is

t =

1

15. 4 × 10 −^3

ln

(

2. 0

0. 5

)

≈ 90 days.

1.3.C CompositeRadionuclides

A problem often encountered in radioactivity measurements is that of determining
the activity of individual elements in a composite material. A composite material
is the one that contains more than one radioisotope at the same time. Most of the
radioactive materials found in nature are composite.
Let us suppose we have a sample that contains two isotopes having very different
half lives. Intuitively thinking, we can say that the semilogarithmic plot of activity