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11 Effective, biological and physical decay

By Dr. Panyi György

11.1 The formula, definitions of the parameters

The number of undecayed nuclei in a living organism can decrease in two independent ways:

due to physical decay, which means, that a nuclear transmutation occurs resulting radioactive

emission, and

due to biological decay, which means that the radioactive nucleus leaves the system

undecayed due to different secretion pathways (e.g. urine, exhalation etc.). Thus, biological

decay does not involve nuclear transmutation, however, the number of undecayed nuclei

within the biological system decreases during this process as well, hence it is called ”decay”.

It is of great importance, when radiopharmaceuticals are administrated into the body for

diagnostic imaging purposes. In order to gain a diagnostic image, an adequate activity is

needed (physical decays/s) in the body. This activity is estimated by considering ”the losses”

as well, resulted by such processes during which the isotope leaves the body undecayed due to

biological decay, (or accumulates in the bladder, for instance). (Undecayed nuclei leaving the

body do not produce the emission required for imaging.)

The following variables can be defined:

physical decay constant (phys): the probability of physical decay per unit time (s-1).

biological decay constant (biol): the probability that an undecayed nucleus leaves a living

system per unit time (s-1).

effective decay constant (eff): the probability, that an undecayed nucleus decayes physically

in the body, OR leaves the system per unit time (s-1).

Since the processes are independent of each other, the decay rate is determined by the

combination of the decays:

𝑑𝑁

𝑑𝑡= −𝜆𝑒𝑓𝑓𝑁 = −(𝜆𝑝ℎ𝑦𝑠 + 𝜆𝑏𝑖𝑜𝑙)𝑁

𝜆𝑒𝑓𝑓 = 𝜆𝑝ℎ𝑦𝑠 + 𝜆𝑏𝑖𝑜𝑙

The solution of the differential equation is the well-known version of the law of radioactive

decay:

𝑁 = 𝑁0𝑒−𝜆𝑒𝑓𝑓𝑡 = 𝑁0𝑒

−(𝜆𝑝ℎ𝑦𝑠+𝜆𝑏𝑖𝑜𝑙)𝑡

In a radioactive preparation, the number of undecayed nuclei (N) exponentially decreases with

time. The decay constant as proportion factor eff can be found in the power.

N0: number of undecayed nuclei at t=0

N: number of undecayed nuclei at the time of investigation

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eff: effective decay constant

t: time

Quantities characterizing the radioactive decay:

Effective half-life of radioactive nuclei (Teff): effective half life gives the time during which

the initial activity of a given type of radioactive nucleus decreases to half of its original value

either by physical decay or metabolism, i.e. the time, where N=N0/2. Its unit is time (s, day,

year).

Physical half-life of radioactive nuclei (Tphys): physical half life is the time period required

to reduce the number of undecayed nuclei to exactly one half its original value due solely to

radioactive decay. Its unit is time (s, day, year)

Biological half-life of radioactive nuclei (Tbiol): biological half life is the time period during

which half of the initial quantity of the radioactive isotope leaves the living system undecayed

due to metabolism, secretion or excretion. Its unit is time (s, day, year)

Knowing the relationship between the decay constant and the half-life:

11.2 Graphical representation of relationships, plotting of relevant

parameters

11.2.1 Linear plot, undecayed nuclei in the body as a function of time

Given N0=4400, Teff=0.67 h, Tbiol=2 h and Tphys=1 h one can construct the graph displaying

the relationship between the number of undecayed nuclei as a function of time for the two

separate processes (physical and biological decay, theoretical) along with the effective decay

(practically happening in the patient).

1/21/2

eff phys biol

eff phys bioleff phys biol

l n 2ln 2 T

T

1 1 1ln 2 ln 2 ln 2

T T TT T T

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Figure 1. Physical, biological and effective decays on linear scale

Key elements of the plotting:

Details of plotting can be found in the previous chapter: The Law of Radioactive Decay

(The first fixed point: at t=0, the number of undecayed nuclei is N0=4400. That is, at t=0, the

value of the function is 4400.

The second fixed point is the value of the function, when t=∞. At this time N~0, since t=∞ is

in the power of the equation N=N0e-∞ (we have to recognize, that e-∞ = (1/ e∞)~0), i.e. if t

is very high, the curve of the function approaches the abscissa (”x axis”).

The last important fixed point of the plotting is that we order the value of N=N0/2=2200 to the

respective half-lives, i.e. in our case:

Tbiol=2 h (the value of half-life, if only biological decay occurred)

Tphys=1 h (the value of half-life, if only physical decay occurred)

Teff=0.67 h (the real value of half-life, since both physical and biological decays occur at the

same time, in an independent manner)

We connect the 3 fixed points mentioned above with such a curve, which STARTS from

N=4400 at t=0, passes through N=2200 corresponding to the half-life, and approaches the

time axis at t=∞.

The problem can be solved by making a semi-logarithmic plot and by plotting ln(N/N0), see

”The Law of Radioactive Decay”.

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11.3 Exercises with solutions:

11.3.1 Exercise

The graph below shows the number of undecayed nuclei (N) as a function of time regarding a

given radioactive decay. Suppose that the red curve represents the physical decay of the

nuclei. Draw another curve, representing the effective decay. Based on the two curves,

determine the half-life of the biological decay.

Figure 2. Illustration for Excercise 11.3.1.

Solution:

Figure 3. Solution for Exercise 11.3.1.

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1, The effective decay is always faster, than the physical decay, thus, we have to draw such a

curve (green color), which is steeper, than the curve of the physical decay.

2, Determination of biological half-life: We have to determine the physical and effective half-

lives graphically. We find and label the value of N=N0/2 t (N=1500) (blue dashed line parallel

to the time axis), then by means of its intersection with the decay curve (red line), using a

vertical dashed dropline, we mark out the physical half-life on the time axis, which is

approximately Tphys=200 min. We proceed similarly, when finding the effective half-life,

which is approximately Teff=125 min.

Knowing Teff and Tphys, Tbiol is calculable:

eff phys biol biol eff phys

biol

1biol

1 1 1 1 1 1 1 1

T T T T T T 125 200

1 1 10.003

T 125 200

T 0.003 333.3min

11.3.2 Exercise

The physical half-life of a given isotope is Tphys= 6 hours. We conjugate the nucleus to a

material (material X), which has a biological half-life of Tbiol=4 hours in a certain patient.

What is the effective half-life of the given radiopharmaceutical (isotope conjugated to

material X) in the patient, if we suppose that the conjugation does not alter the biological half-

life of material X? After how long time will the number of undecayed nuclei decrease to ¼

and 1/8 of the original value?

Solution:

phys

biol

eff phys biol eff

1eff

T 6h

T 4h

1 1 1 1 1 10.416

T T T T 6 4

T 0.416 2.4h

Since the half-life is 2.4 hours, thus, the number of undecayed nuclei decreases to 1/2, ¼ and

1/8 of the initial number of nuclei after 2.4, 4.8 and 7.2 hours, respectively (1,2 and 3 half-

lives).

11.3.3 Exercise

The radiopharmaceutical mentioned in the exercise above is applied in a bone scintigraphy

diagnostic imaging. The radiopharmaceutical with an activity of 740 MBq is administrated

into the body intravenously. What is the longest time that can be passed between the

administration and the gamma camera imaging, given that the gamma camera requires at least

an activity of 200 MBq?

Solution:

part 1: determination of the number of nuclei that has be present at minimum at the time of

the examination. The activity is determined by the physical decay (only the radioactive decay

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produces measurable emission). We find the N0 value at t=0, and we find the N1 value at the

unknown t1 time:

phys 0

66 6phys 13

0phys

66 6phys 12

1phys

dNt 0, N

dt

740 10 T740 10 740 10 6 3600N 2.3 10

ln 2 ln 2

200 10 T200 10 200 10 6 3600N 6.23 10

ln 2 ln 2

part 2: determination of the longest time until the examination. The number of undecayed

nuclei decreases from N0 to N1 based on the kinetics of the effective decay, the corresponding

time can be described by the law of radioactive decay:

eff

130 0

eff 121 1

eff

t1 0

0eff

1

N N 2.3 10ln ln T ln 2.4 3600

N N 6.23 10ln 2 ln 2

N N e

Nln t

N

t 16280 s 4.52 h

That is, the examination must be performed in 4.52 hours.

(Based on the previous exercises, the problem can be solved in a simpler way by using the

activities (A) directly for the calculations:

eff

60 0

eff 61 1

eff

t1 0

0eff

1

A A 740 10ln ln T ln 2.4 3600

A A 200 10ln 2 ln 2

A A e

Aln t

A

t 16308 s 4.53 h

11.3.4 Exercise

The physical and biological half-lives of a conjugated isotope are known: Tphys=6 hours and

Tbiol=3.5 hours. Determine the effective decay constant.

Solution 1

eff phys biol

eff

1eff

5 1eff

eff

1 1 1 1 10.4524

T T T 6 3.5

10.4524

T

T 0.4524 2.21h 7956 s

ln 28.71 10 s

T

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Solution 2

5 1phys

phys

5 1biol

biol

5 1 5 1 5 1eff phys biol

ln 2 ln 23.2 10 s

T 6 3600

ln 2 ln 25.5 10 s

T 3.5 3600

3.2 10 s 5.5 10 s 8.7 10 s

11.3.5 Exercise

The half-life of the 123I isotope is 13.22 hours. This isotope was administrated into a patient,

and the table below shows the number of undecayed nuclei in the body as a function of time.

Plot the physical, biological and effective decays in the same coordinate system based on the

given information. Based on the curves, determine the effective and biological half-lives.

Time (h) 0 5 10 15 20 40

~N 100,000 52,000 27,000 14,000 7 250 500

Solution:

1, Knowing the physical half-life, we draw the curve characteristic for the decay (red curve).

See chapter 11.2.1 and the law of radioactive decay chapters for constructing the curve.

2, We plot the number of undecayed nuclei as a function of time based on data given in the

table. These points are characteristic for the effective decay (green color).

Figure 4. Solution for Exercise 11.3.5.

3, We connect these effective decay data points, and we determine the effective half-life,

which is approximately : Teff~5h. For drawing the curve see chapter 11.2.1 and the Law of

Radioactive Decay chapters.

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Figure 5. Solution for Exercise 11.3.5. (part 2)

4, We calculate the biological half-life and draw the biological decay curve. For drawing the

curve see chapter 11.2.1 and the Law of Radioactive Decay chapters.

eff phys biol biol eff phys

biol

1biol

1 1 1 1 1 1 1 1

T T T T T T 5 13.2

1 1 10.124h

T 5 13.2

T 0.124 8.06h 8h

Figure 6. Solution for Exercise 11.3.5.(part 3)

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11.4 Exercise

1. If the physical half-life of a radioactive isotope is 18h, and its biological half-life is 9h,

when will the number of radioactive nuclei decrease to ¼ of the original value?

(a) 6h

(b) 27h

(c) 12h

2. If the physical half-life of a radioactive isotope is 18h, and its biological half-life is 9h,

what is the value of the effective decay constant (expressed in s-1 )?

(a) 0.0577 s-1

(b) 3.210-5 s-1

(c) 0.037 s-1

3. The physical half-life of a radioactive isotope is 16h, the biological half-life of a

conjugated radiopharmaceutical is 9.5 hours. The activity of the isotope administrated

to a patient’s body was 10 MBq. What was the number of undecayed nuclei at the time

of the radiopharmaceutical administration?

(a) 8.31011

(b) 2.3108

(c) 3.1251011

4. The physical half-life of a radioactive isotope is 16h, the biological half-life of a

conjugated radiopharmaceutical is 9.5 hours. What is the effective half-life of the

radiopharmaceutical?

(a) ~6h

(b) 25 h

(c) 0.167 h

5. If we administrate 9×1011 undecayed nuclei into a living organism, with a physical

half-life of 11.2 hours, and biological half-life of 6.9 hours, what will be the activity

after 15 hours?

(a) 0.51 MBq

(b) 1.35 MBq

(c) 3.55 MBq

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6. The red curve on the graph represents the physical decay of an isotope. Which one of

the green curves DOES NOT indicate the effective decay?

(a)

(b)

(c)

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7. The red curve on the graph represents the physical decay of an isotope. Which one of

the green curves DOES NOT indicate the effective decay?

(a)

(b)

(c)

8. 123I isotope with an activity of 0.3 MBq is needed for a diagnostic examination (0.3

MBq=0.3× 106 decays/s). What is the required number of undecayed nuclei, if the

physical half-life of 123I is 13.22h and its biological half life is 6.9h?

(a) 2× 1010

(b) 5.72× 106

(c) 8× 107

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9. The physical half-life of a radioactive isotope is exactly twice of its biological half-

life. What is the relationship between the physical and effective half-lives?

(a) 3/2Teff=Tphys

(b) 3Teff=Tphys

(c) Teff=3/2 Tphys

10. How much time is needed until the number of undecayed nuclei administrated into a

body decreases to 6.25% of the original value, if the biological half-life is 6h, and the

physical half-life is 9h?

(a) 14.4 h

(b) 1.11 h

(c) 57.6 h