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.