Physics and Engineering of Radiation Detection

1.3. Radioactivity and Radioactive Decay 17

This straight line represents the activity of the long lived component in the

sample. Its slope gives the decay constant of the long lived component, which

can then be used to obtain the half life. Hence we get

λ 1 =3. 28 × 10 −^3 min−^1

⇒T 1 / 2 , 1 =

0. 693

λ 1

=

0. 693

3. 28 × 10 −^3

= 211.3days.

To obtain the decay constant of the short lived component, we extrapolate

the straight line obtained for the long lived component up tot=0andthen

subtract it from the observed data (see Fig.1.3.4). The straight line thus

obtained is given by

ln(A)=− 3. 55 × 10 −^2 t+7. 53.

The slope of this line gives the decay constant of the second isotope, which

can then be used to determine its half life. Hence we have

λ 2 =3. 55 × 10 −^2 min−^1

⇒T 1 / 2 , 2 =

0. 693

λ 1

=

0. 693

3. 55 × 10 −^2

=19.5days.

Time (min)

0 50 100 150 200 250 300

Counts

102

103

Figure 1.3.4: Determination of decay parameters of two

nuclides from observedeffectiveactivity. The actual data

are represented by (*). The solid and dashed lines represent

the long lived and short lived components respectively.