Ahmed S.N. Physics and Engineering of Radiation Detection

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16 Chapter 1. Properties and Sources of Radiation

versus time in such a case will deviate from a straight line of single isotopes. The

best way to understand this is by assuming that the composite material has one

eﬀective decay constant. But this decay constant will have time dependence since as

time passes the sample runs out of the short lived isotope. Hence equations 1.3.20

and 1.3.21 will not be linear any more.

Figure 1.3.3 shows the activity plot of a composite radioactive material. Since

we know that each individual isotope should in fact yield a straight line therefore

we can extrapolate the linear portion of the graph backwards to get the straight

line for the isotope with longer half life. We can do this because the linear portion

shows that the shorter lived isotope has fully decayed and the sample now essentially

contains only one radioactive isotope. Then the straight line for the other isotope

can be determined by subtracting the total activity from the activity of the long

lived component.

ln(A)

Tim

Slow decaying component

Fast decaying component

Measured activity

Figure 1.3.3: Experimental de-

termination of decay constants

of two nuclides in a composite

decaying material.

Example:

The following table gives the measured activity in counts of a composite ra-

dioactive sample with respect to time. Assuming that the sample contains two

radioactive isotopes, compute their decay constants and half lives.

t (min) 0 30 60 90 120 150 180 210 240 270 300

A (cts/min) 2163 902 455 298 225 183 162 145 133 120 110

Solution:

Following the procedure outlined in this section, we plot the activity as a

function of time on a semilogarithmic graph (see Fig.1.3.4). It is apparent from

theplotthataftert = 120 minutes ln(A) varies linearly with time. Using least

square ﬁtting algorithm we ﬁt a straight line through points between t = 150

and t = 300 minutes. The equation is found to be

ln(A)=−3.28 × 10

−3

t +5.68.

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

=3.28 × 10

−3

min

−1

⇒ T

1/2,1

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 to t =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

=3.55 × 10

−2

min

−1

⇒ T

1/2,2

0.693

3.55 × 10

−2

=19.5days.

Time (min)

0 50 100 150 200 250 300

Counts

Figure 1.3.4: Determination of decay parameters of two

nuclides from observed eﬀective activity. The actual data

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

the long lived and short lived components respectively.

18 Chapter 1. Properties and Sources of Radiation

1.3.D Radioactive Chain

We saw earlier that when a radionuclide decays, it may change into another element

or another isotope. This new daughter radionuclide may as well be unstable and

radioactive. The decay mode and half life of the daughter may also be diﬀerent

from the parent. Let us see how our radioactive decay equations can be modiﬁed for

such a situation.

I will start with a sample composed of a parent and a daughter radionuclide.

There will be two processes happening at the same time: production of daughter

(or decay of parent) and decay of daughter. The net rate of decay of the daughter

will then be the diﬀerence of these two rates, that is

= λ

− λ

, (1.3.24)

where subscripts P and D represent parent and daughter respectively.

Using N

= N

−λ

, this equation can be written as

+ λ

− λ

−λ

=0. (1.3.25)

Solution of this ﬁrst order linear diﬀerential equation is

− λ



−λ

− e

−λ



+ N

−λ

. (1.3.26)

Here N

and N

are the initial number of parent and daughter nuclides respec-

tively. In terms of activity A(= λN), the above solution can be written as

− λ



−λ

− e

−λ



+ A

−λ

. (1.3.27)

Equations 1.3.26 and 1.3.27 have decay as well as growth components, as one would

expect. It is apparent from this equation that the way a particular material decays

depends on the half lives (or decay constants) of both the parent and the daughter

nuclides. Let us now use equation 1.3.27 to see how the activity of a freshly pre-

pared radioactive sample would change with time. In such a material, the initial

concentration and activity of daughter nuclide will be zero N

=0,A

=0. This

condition reduces equation 1.3.26 to

− λ



−λ

− e

−λ



. (1.3.28)

The ﬁrst term in parenthesis on the right side of this equation signiﬁes the buildup

of daughter due to decay of parent while the second term represents the decay of

daughter. This implies that the activity of the daughter increases with time and,

after reaching a maximum, ultimately decreases (see ﬁgure 1.3.5). This point of

maximum daughter activity t

max

can be easily determined by requiring

=0.

Applying this condition to equation 1.3.28 gives

max

ln(λ

/λ

)

− λ

. (1.3.29)

1.3. Radioactivity and Radioactive Decay 19

Activity

max

Tim

Figure 1.3.5: Typical parent

and daughter nuclide activities.

Example:

Derive the relation for the time behavior of buildup of a stable nuclide from

a radioactive element.

Solution:

Assuming the initial concentration of daughter to be zero (N

= 0), equation

1.3.26 can be used to determine the concentration of daughter nuclide at time

− λ



−λ

− e

−λ



Since the daughter nuclide is stable therefore we can substitute λ

=0in

the above equation to get the required relation.

= N



1 − e

−λ



In most cases the radioactive decay process does not stop at the decay of the

daughter nuclide as depicted by equation 1.3.26. Instead the nuclides continue to

decay into other unstable nuclides until a stable state is reached. Assuming that the

initial concentrations of all the nuclides except for the parent is zero, equation 1.3.28

can be generalized for a material that undergoes several decays. The generalization

was ﬁrst done by Bateman in 1910 (8). The Bateman equation for the concentration

of i

radionuclide is

(t)=λ

.....λ

d(i−1)



j=1

−λ



k=1,k=j

(λ

− λ

)

, (1.3.30)

provided N

=0fori>1. In terms of activity the Bateman equation can be

written as

(t)=λ

.....λ



j=1

−λ



k=1,k=j

(λ

− λ

)

. (1.3.31)

20 Chapter 1. Properties and Sources of Radiation

Example:

A50µCi radioactive sample of pure

222

Rn goes through the following series

of decays.

222

Rn (T

1/2

=3.82 days ) →

218

Po (T

1/2

=3.05 min ) →

214

Pb (T

1/2

=26.8

min ) →

214

Bi (T

1/2

=19.7min)

Compute the activity of its decay products after 3 hours.

Solution:

Activity of

222

The decay constant of

222

Rn can be calculated from its half life as follows.

ln(2)

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

−λ

=50



−1.26×10

−4

×3×60



=48.88 µCi

Activity of

218

The decay constant of

218

Po is

ln(2)

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 use i = 2 in Bateman equation 1.3.31 to get

= λ



−λ

(λ

− λ

)

−λ

(λ

− λ

)



=0.227 × 50



−1.26×10

−4

×3×60

(0.227 − 1.26 ×10

−4

)

−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.

1.3. Radioactivity and Radioactive Decay 21

Activity of

214

The decay constant of

214

Pb is

ln(2)

1/2

0.693

26.8

=0.0258 min

−1

To calculate the activity of this isotope of lead after 3 hours we use i =3in

Bateman equation 1.3.31.

= λ



−λ

(λ

− λ

)(λ

− λ

)

−λ

(λ

− λ

)(λ

− λ

)





−λ

(λ

− λ

)(λ

− λ

)



Due to high decay constants of

218

Po and

214

Pb the second and third terms

on right hand side can be neglected. Hence

≈ λ



−λ

(λ

− λ

)(λ

− λ

)



=0.227 × 0.0258 ×50



−1.26×10

−4

×3×60

(0.227 −1.26 ×10

−4

)(0.0258 − 1.26 × 10

−4

)



=49.14 µCi.

Activity of

214

The decay constant of

214

Bi is

ln(2)

1/2

0.693

19.7

=0.0352 min

−1

The Bateman’s equation for i = 4 will also contain negligible exponential

terms as we saw in the previous case. Hence we can approximate the activity

214

Bi after 3 hours by

≈ λ



−λ

(λ

− λ

)(λ

− λ

)(λ

−λ

)



=49.32 µCi.

22 Chapter 1. Properties and Sources of Radiation

1.3.E Decay Equilibrium

Depending on the diﬀerence between the decay constants of parent and daughter

nuclides it is possible that after some time their activities reach a state of equilib-

rium. Essentially there are three diﬀerent scenarios leading to diﬀerent long term

states of the radioactive material. These are termed as secular equilibrium, tran-

sient equilibrium,andno equilibrium states. For the discussion in this section we

will assume a radioactive material that has a parent and a daughter only. However

the assertions will also be valid for materials that go through a number of decays.

E.1 Secular Equilibrium

If the activity of parent becomes equal to that of the daughter, the two nuclides are

said to be in secular equilibrium. This happens if the half life of parent is much

greater than that of the daughter, i.e.

1/2

 T

1/2

or λ

 λ

Let us see if we can derive the condition of equal activity from equation 1.3.28. It

is apparent that if λ

>λ

then as t →∞

−λ

 e

−λ

and hence we can neglect the second term on right side of equation 1.3.28. The

daughter activity in this case is given approximately by



− λ

−λ

(1.3.32)

− λ

 1 −

. (1.3.33)

As λ

 λ

we can neglect second term on right hand side of this equation.

Hence

 A

This shows that if the half life of parent is far greater than that of the daughter then

the material eventually reaches a state of secular equilibrium in which the activities

of parent and daughter are almost equal. The behavior of such a material with

respect to time is depicted in Fig.1.3.6.

An example of a material that reaches secular equilibrium is

237

Np. Neptunium-

237 decays into protactinium-233 through α-decay with a half life of about 2.14×10

years. Protactinium-233 undergoes β-decay with a half life of about 27 days.

Example:

How long would it take for protactanium-233 to reach secular equilibrium

with its parent neptunium-237?

1.3. Radioactivity and Radioactive Decay 23

Activity

1/2

Tim

Figure 1.3.6: Activities of par-

ent and daughter nuclides as a

function of time for a material

that eventually reaches the state

of secular equilibrium. The half

life of parent in such a material

is so large that it can be consid-

ered stable.

Solution:

Since the half life of the parent is much larger than that of the daughter,

therefore we can safely assume that at the state of secular equilibrium the

activity of the daughter will be nearly equal to the initial activity of the

parent, that is.

 A

Substitution of this equality in equation 1.3.32 gives

− λ

−λ

 1

⇒ t 



− λ



The decay constants of the two materials are

ln(2)

1/2

ln(2)

2.14 × 10

× 365

=8.87 × 10

−10

day

−1

ln(2)

1/2

ln(2)

=2.57 × 10

−2

day

−1

Hence the required time is given by

t 

8.87 × 10

−10



2.57 × 10

−2

2.57 × 10

−2

− 8.87 × 10

−10



=38.9days

24 Chapter 1. Properties and Sources of Radiation

Interestingly enough, this is exactly the mean life of protactanium-233 and

it shows that after about one mean life of the daughter its activity becomes

approximately equal to that of the parent.

E.2 Transient Equilibrium

The parent and daughter nuclides can also exist in a transient state of equilibrium in

which their activities are not equal but diﬀer by a constant fraction. This happens

when the half life of the parent is only slightly higher than that of the daughter, i.e.

1/2

or λ

<λ

The approximate activity 1.3.33 derived earlier is valid in this situation as well.

 1 −

However now we can not neglect the second term on the right side as we did in the

case of secular equilibrium. In this case the equation depicts that the ratio of parent

to daughter activity is a constant determined by the ratio of parent to daughter

decay constant. Figure 1.3.5 shows the typical behavior of such a material. A

common example of transient equilibrium decay is the decay of Pb

212

into Bi

212

E.3 No Equilibrium

If the half life of parent is less than the half life of daughter, i.e.

1/2

or λ

>λ

then the activity due to parent nuclide will diminish quickly as it decays into the

daughter. Consequently the net activity will be solely determined by the activity of

the daughter. Figure 1.3.7 depicts this behavior graphically.

1.3.F Branching Ratio

In the preceding sections we did not make any assumption with regard to whether

there was a single mode or multiple modes of decay of the nuclides. In fact the

majority of the nuclides actually decay through a number of modes simultaneously

with diﬀerent decay constants. Branching ratio is a term that is used to characterize

the probability of decay through a mode with respect to all other modes. For example

if a nuclide decays through α and γ modes with branching ratios of 0.8 and 0.2, it

would imply that α-particle is emitted in 80% of decays while photons are emitted

in 20% of decays. The total decay constant λ

d,t

of such a nuclide having N decay

modes is obtained by simply adding the individual decay constants.

d,t



i=1

d,i

(1.3.34)

Here λ

d,i

represents the decay constant of the ith mode for a material that decays

through a total of n modes. The total decay constant can be used to determine the

1.3. Radioactivity and Radioactive Decay 25

1/2

Activity

Tim

Figure 1.3.7: Activities of par-

ent and daughter nuclides as a

function of time for a material

that never reaches the state of

equilibrium. The parent in such

a material is shorter lived than

the daughter.

eﬀective activity and other related quantities. The expressions for the eﬀective half

and mean lives can be obtained by substituting T

1/2,i

=0.693/λ

d,i

and τ

=1/λ

d,i

for the ith decay mode in the above equation. This gives

1/2.e



i=1

1/2,i

(1.3.35)

and



i=1

, (1.3.36)

where T

1/2,e

and τ

represent the eﬀective half and mean lives respectively.

1.3.G Units of Radioactivity

Since the most natural way to measure activity of a material is to see how many

disintegrations per unit time it is going through, therefore the units of activity are

deﬁned in terms of disintegrations per second. For example 1 Becquerel corresponds

to 1 disintegration per second and 1 Curie is equivalent to 3.7 ×10

disintegrations

per second. Curie is a much bigger unit than Becquerel and is therefore more

commonly used. However for most practical sources used in laboratories, Curie is

too big. Therefore its subunits of milli-Curie and micro-Curie are more commonly

found in literature. The subunits of Curie and interconversion factors of Curie and

Becquerel are given below.