1.3. Radioactivity and Radioactive Decay 21
Activity of^21482 Pb
The decay constant of^21482 Pbisλd 3 =ln(2)
T 1 / 2=
0. 693
26. 8
=0.0258 min−^1.To calculate the activity of this isotope of lead after 3 hours we usei=3in
Bateman equation 1.3.31.A 3 =λd 2 λd 3 A 01[
e−λd^1 t
(λd 2 −λd 1 )(λd 3 −λd 1 )+
e−λd^2 t
(λd 1 −λd 2 )(λd 3 −λd 2 )]
+
λd 2 λd 3 A 01[
e−λd^3 t
(λd 1 −λd 3 )(λd 2 −λd 3 )]
Due to high decay constants of^21884 Poand^21482 Pbthe second and third terms
on right hand side can be neglected. HenceA 3 ≈ λd 2 λd 3 A 01[
e−λd^1 t
(λd 2 −λd 1 )(λd 3 −λd 1 )]
=0. 227 × 0. 0258 × 50
[
e−^1.^26 ×^10− (^4) × 3 × 60
(0. 227 − 1. 26 × 10 −^4 )(0. 0258 − 1. 26 × 10 −^4 )
]
=49. 14 μCi.Activity of^21483 Bi
The decay constant of^21483 Biisλd 4 =ln(2)
T 1 / 2=
0. 693
19. 7
=0.0352 min−^1.The Bateman’s equation fori = 4 will also contain negligible exponential
terms as we saw in the previous case. Hence we can approximate the activity
of^21483 Biafter 3 hours byA 4 ≈ λd 2 λd 3 λd 4 A 01[
e−λd^1 t
(λd 2 −λd 1 )(λd 3 −λd 1 )(λd 4 −λd 1 )]
=49. 32 μCi.