Physics and Engineering of Radiation Detection

20 Chapter 1. Properties and Sources of Radiation

Example:

A50μCiradioactive sample of pure^22288 Rngoes through the following series

of decays.

222

86 Rn(T^1 /^2 =3.82 days )→

218

84 Po(T^1 /^2 =3.05 min )→

214

82 Pb(T^1 /^2 =26.^8

min )→^21483 Bi(T 1 / 2 =19.7min)

Compute the activity of its decay products after 3 hours.

Solution:

Activity of^22286 Rn

The decay constant of^22286 Rncan be calculated from its half life as follows.

λd 1 =

ln(2)

T 1 / 2

=

0. 693

3. 82 × 24 × 60

=1. 26 × 10 −^4 min−^1

Since we have a pure sample of radon-222 therefore its activity after 3 hours

can be calculated from equation 1.3.16.

A 1 = A 01 e−λd^1 t

=50

[

e−^1.^26 ×^10

− (^4) × 3 × 60 ]
=48. 88 μCi
Activity of^21884 Po
The decay constant of^21884 Pois
λd 2 =
ln(2)
T 1 / 2

=

0. 693

3. 05

=0.227 min−^1.

Since polonium-218 is the first daughter down the radioactive chain of radon-

222, we usei= 2 in Bateman equation 1.3.31 to get

A 2 = λd 2 A 01

[

e−λd^1 t

(λd 2 −λd 1 )

+

e−λd^2 t

(λd 1 −λd 2 )

]

=0. 227 × 50

[

e−^1.^26 ×^10

− (^4) × 3 × 60
(0. 227 − 1. 26 × 10 −^4 )

+

e−^0.^227 ×^3 ×^60

(1. 26 × 10 −^4 − 0 .227)

]

=48. 9 μCi

A point worth noting here is that the second term in the parenthesis on the

right side of the above equation is negligible as compared to the first term

and could have safely been omitted from the calculations.