0481 - H201 - Health Physics Technology - 07 - Mean Life. · ¾Discuss the concept of effective half-life and derive it mathematically ¾Discuss significance of effective half-life

Chapter 07HRTDHuman ResourcesTraining & Development

- Slide 1 -H-201 - Health Physics Technology

Mean Life

Discuss the concept of effective half-life and derive it mathematically

Discuss significance of effective half-life

Objectives


Calculate the effective half life, the radiological half life, or the biological half life given information about the other two quantities

MEAN LIFE


MEAN LIFE

Understand and discuss the significance of the mean life

Calculate the mean life given the half-life of a given

Objectives


g gradioisotope

Use the mean life to calculate external dose

Explain the use of mean life in other Applications

Radioactive Decay

The area under the curve is Activity x Time (d/t) x t = dwhich is the total number of disintegrations

Activity (A)

μCi


(But how do we determine the area under the curve?)

or

disintegration

time

time (t)

ExampleVehicle Traveling at a Constant Speed

Speed (s)50


The area under the curve is speed x time or(50 mi/hr) x 1 hr = 50 miles

mph

or

miles

hour

time (hours) 1

Speed (s)

mph

50

The area under the curve is ( d ti )/2

ExampleDecelerating Vehicle


mph

or

miles

hour

time (hours) 1

(speed x time)/2 or(50 mi/hr x 1 hr)/2= 25 miles

To obtain the area under the decay curve which is not a straight line, we must integrate the decay equation

A = Ao e - λt ∫∞

A dt = Ao e - λt dt

Area Under the Decay Curve

∫∞


∫0

= Ao e - λt dt

= Ao

0

∞e - λt

-λ

∫0

∫0

∞

Substituting ∞ and 0 for t

= Aoe - λ(∞)

-λ- e - λ(0)

- λ



= Ao- λ

- 1- λ

0

= +Aoλ

0 1=

Ao

λ

So the total area under the decay curve is just the initial activity (Ao) divided by the decay constant. The activity has units of d/t and λ has units of 1/t so the units are (d/t) / (1/t) = (d/t) x t = d which is what we wanted.



How does the half life (T) fit in?

Well we know that λ = ln(2)/T = 0.693/T

So Ao

λ=

0.693

T= 1.44 Ao T = Ao Tm

Ao

where Tm = mean life = 1.44 T

So going back a few slides we found that the area under the decay curve {which represented all the initial activity (Ao) converted to disintegrations} was equal to

A

Mean Life


λAoArea Under the Decay Curve = = 1.44 T Ao = Ao Tm

Where

Tm = 1.44 Tr for external exposure

The area under the decay curve is Ao/λ

Tm = 1.44 T½

Activity (A)

μCi

or ½Ao

Ao

Mean Life


T½ = Tr for external

(for external, the activity can be replaced by the dose rate)

TmT½

or

disintegration

time

time (t)

o

The area under the rectangle (Ao x Tm) is identical to the area under the exponential decay curve since Tm = 1/λ

Activity (A)

μCi

Ao

The portion of the curve in green (///) is common to both. Therefore, the remaining portion of the curve in blue (|||) must

Mean Life


or

disintegration

time

time (t)Tm

½Ao

curve in blue (|||) must be equal to the portion of the rectangle in purple (Ξ).

One might think that it’s impossible for the finite purple area under the rectangle to be equal to the blue area under the tail of the decay curve since the decay curve is exponential indicating that it goes out to ∞. But in reality, we start with a finite number of atoms and over some

Mean Life


we start with a finite number of atoms and over some period of time eventually all the atoms will transform so that the rectangle does in fact represent all the atoms disintegrating over a hypothetical finite time Tm.

Remember the activity equation A = λNThis indicates that total number of atoms N = A/λ = ATm

To determine the dose over a specific time period (e.g., 50 years), you can use the following equation to determine the actual number of transformations which occur from t = 0 to 50 years (or any other time interval t = 0 to t).

Dose Over t Years


At Tm = Ao Tm(1 - e ) -

AtTm = Ao Tm - Aoe Tm

-0.693 t

Tr

Simplified -0.693 t

Tr-0.693 t

Tr

At Tm = Ao Tm - Ao e Tm

0.693 t

Tr

-

Dose Over “t” Years


Where Ao e is just the activity left at time t (A1)

So the equation gives us the number of transformations from 0 to ∞ minus the transformations from “t” to ∞ which yields the number of transformation from 0 to “t”

0.693 t

Tr

-

If t = 0, AtT = 0 (no transformations have occurred)

At Tm = Ao Tm - Ao e Tm

0.693 t

Tr

-



If t 0, AtTm 0 (no transformations have occurred)

if t = ∞, AtTm = AoTm (all atoms have transformed)


Remember, you CAN use the mean life (AoTm for t = ∞) to approximate the CDE over 50 years, IF the effective half life (Te) is relatively short. In that case all the material decays within 50 years.

(1 - e )0.693 t

Tr

- is essentially a correction factor for the case where t < ∞

At Tm = Ao Tm (1 - e )0.693 t

Tr

-Starting with:



For example, say Tr is 5 years.

How many transformations occur after 1 year?

A1 Tm = Ao Tm(1 - e ) = AoTm (0.13) 13% 0.693 x 1

5-

After t = 1 year, only 13% of the atoms have transformed compared to 100% at t = ∞.

Assume that the radiological half life is 5 years (60Co).

We can say that if an individual remained near the “source” for a “time” equal to Tm (1.44 x 5 years = 7.2 years) then we can determine all of the dose received

Mean Life Concept


from the source (until the last atom transforms).

What we are actually saying is NOT that the individual would receive all of the dose in 7.2 years, but that we can CALCULATE all of the dose he/she would receive by using Tm = 7.2 years and assuming the source does not decay during that time.

He/she would not actually receive all the dose until the last atom decayed which might be 100 years in the future (assuming he/she lived that long).

In fact, if you use the exponential equation from the

Mean Life Concept


previous slide you will find that the individual will actually receive 63% of the total dose after 7.2 years and 99.9% of the total dose after 50 years.

But we can calculate the total dose received simply by using Tm = 7.2 years times the initial activity of the source.

So we can’t say WHEN the total dose will actually be received, but we can calculate the AMOUNT of the dose by making the assumption that he/she is exposed to a “constant activity source” for 7.2 years (i.e., the initial activity (curies) does not decrease but stays constant

Mean Life Concept


activity (curies) does not decrease but stays constant until the last atom decays and then the activity goes abruptly to zero).

At Tm = Ao Tm(e - e )0.693 t2

Te

-0.693 t1

Tr

-

General Form of theMean Life Equation

If we want to determine the dose received between time = t1 and time = t2, we can use this equation:


At Tm = Ao Tm(1 - e )0.693 t

Tr

-This general form reduces to:

when t1 = 0 and t2 = t

General Form of theMean Life Equation

The general form reduces to:

when t = 0 and t = ∞

At Tm = Ao Tm


when t1 = 0 and t2 = ∞

So, if you are starting at time zero, you can always use the simpler forms. However, if you are starting at some time other than zero, you’ll need to use the general form.

For example, if you want to calculate the dose starting on day 3 out to ∞, use the general form of the equation or just shift the x-axis to day 3

original Ao

new Ao

Dose From Time t1


Using the decay equation calculate a new Ao and ignore the one given initially

time (t)Tmt=3or new t=0

Sample Problem

Suppose an individual is exposed to a gamma source at a distance of 1 meter for 8 hours per day for 3 days. What is their total dose?

If the source is a long half life radionuclide like 137Cs,


gthe total dose is the dose rate (mR/hr) times 24 hrs since the dose rate is constant over the 3 days.

But what if the material has a short half life so that it decays over the 3 days. In that case the dose rate changes so the problem becomes more complicated.

Sample Problem

Let’s start with a simpler problem. Suppose the half life is short but the individual is exposed continuously for 3 days. What is their total dose?

The total dose from time 0 to ∞ would be:


Drate(0) x Tm = mR (we’re just substituting Drate for Ao)

where

Drate(0) is just Γ x activity/distance2 = mR/hr

Tm = 1.44 T½ = hr (this value never changes)

Sample Problem

But the Drate changes so what would it be after 3 days?

Drate(3) = Drate(0) x e(-0.693x3/T½) (since the activity decays exponentially)

So the dose rate from 3 days to ∞ would be:


Drate(3) x Tm = Drate(0) x e(-0.693x3/T½) x Tm

If we subtract the two equations we get:

Drate(0) x Tm - Drate(0) x e(-0.693x3/T½) x Tm =

Drate(0) x Tm x (1-e (-0.693x3/T½)) (this is the dose from 0 to 3 days)

Sample Problem

Visually what we have is this:

We first calculated the dose from 0 to ∞ (red squiggles) and then subtracted out the dose from 3 to


0 3

subtracted out the dose from 3 to ∞ (blue triangle) which leaves us with the dose from 0 to 3 days.

Sample Problem

Now let’s get back to our original scenario of an individual being exposed only 8 hrs per day. That would look something like this:

We’re looking for the first 8 hrs


0 72

of every 24 hr day. So we calculate the dose for 8 hrs, no dose for 16 hrs etc. That gives us the shaded areas. This looks complicated. Or is it?

824

3248

56

First question - how much of the source is left after any 24 hr period?

Sample Problem

That’s just e(-0.693 x 24/T½)

Second question how much does the source


0 728

2432

4856

That’s just (1-e(-0.693x8/T½))

Second question - how much does the source decay in an 8 hr period?

The beauty of radioactive decay is that a constant fraction decays per unit time

So, in the first 8 hr period we have:

Drate(0) x Tm x (1-e(-0.693x8/T½)) (that’s 0 to ∞ minus 8 to ∞)

Sample Problem

The new starting dose rate is:


0 728

2432

4856

The new starting dose rate is:

Drate(24) = Drate(0) x e(-0.693 x 24/T½)

And we use it to get the second dose interval:

Drate(24) x Tm x (1-e(-0.693x8/T½))

(that’s 24 to ∞ minus 32 to ∞)

You may have noticed a pattern here. The time used to calculate the dose in the 8 hr interval is always the same, namely:

Tm x (1-e(-0.693x8/T½))

Sample Problem


m ( )

when T½ is very very large, the expression reduces to Tm

which is what we would use for 137Cs since the dose rate doesn’t change as time passes

But when T½ is very small, the time interval gets very small - since the material is decaying quickly the time of exposure at any given dose rate is small

Let’s do the problem for 137Cs which has a half life of 30 years and then repeat it for 201Tl which has a half life of 3 days.

Sample Problem

D = (0 33 R-m2/hr-Ci) x (1 Ci)/(1 m)2 = 0 33 R/hr


Drate(0) = (0.33 R-m /hr-Ci) x (1 Ci)/(1 m) = 0.33 R/hr

The two fractions that we need to calculate for 137Cs are:

(1-e(-0.693 x 8/30x365x24)) = 0.000021

e(-0.693 x 24/30x365x24) = 0.999937

Let’s do the math.

Sample Problem

Drate(24) = 0.33 R/hr x 0.999937 = 0.32998 R/hrDrate(48) = 0.32998 R/hr x 0.999937 = 0.32996 R/hr


The time interval over which the dose is accrued is:

Tm x 0.000021 = 1.44 x 30 x 365 x 24 x 0.000021 = 7.95 hr

Because the source is decaying, the 8 hr period appears to be like a 7.95 hr period.

So the doses accrued in the three 8 hr periods is:

Sample Problem

D0 = 0.33 R/hr x 7.95 hr = 2.6235 RD24 = 0.32998 R/hr x 7.95 hr = 2.6233 RD48 = 0.32996 R/hr x 7.95 hr = 2.6232 R


Total = 7.87 R

Which is not much different than if we had assumed the dose rate was constant (0.33 R/hr) and multiplied it by the 24 hrs the individual was exposed

D = 0.33 R/hr x 24 hr = 7.92 R

Now let’s repeat the problem for 201Tl which has a half life of 3 days. Although the Γ for 201Tl is 0.46 compared to 0.33 for 137Cs, let’s assume the two Gamma Constants are the same.

Sample Problem


Drate(0) = (0.33 R-m2/hr-Ci) x (1 Ci)/(1 m)2 = 0.33 R/hr

The two fractions that we need to calculate for 201Tl are:

(1-e(-0.693 x 8/3x24)) = 0.074

e(-0.693 x 24/3x24) = 0.794


Sample Problem




Tm x 0.074 = 1.44 x 3 x 24 x 0.074 = 7.67 hr


Sample Problem



Total = 6.14 R

This is certainly less than what we got when we did the calculation for 137Cs but not as much different as we might have thought given the significant difference between the two half lives (3 days and 10,950 days).

Sample Problem

That’s because they both started off at the same dose rate and most of the dose is delivered in the first few days. If we continued calculating 8 hr doses out to 10 days, the Cs dose would be relatively constant at about 2.62 R/period while the Tl dose is decreasing so that it will


R/period while the Tl dose is decreasing so that it will eventually get to 0 R/period. The maximum total dose would be achieved fairly soon.

Let’s do one more example using 99mTc which has a half life of 6 hrs.

Although the Γ for 99mTc is 0.078 compared to 0.33 for 137Cs, let’s again assume the two Gamma Constants are the same.

Sample Problem


Drate(0) = (0.33 R-m2/hr-Ci) x (1 Ci)/(1 m)2 = 0.33 R/hr

The two fractions that we need to calculate for 99mTc are:

(1-e(-0.693 x 8/6)) = 0.603

e(-0.693 x 24/6) = 0.063


Sample Problem




Tm x 0.603 = 1.44 x 6 x 0.603 = 5.21 hr


Sample Problem



Total = 1.84 R

This is significantly less than what we got for Cs or Tl. The material decays so quickly that even within the first 8 hrs the dose rate decreases dramatically and there’s very little left to produce a dose in the second and third days.

At 6 PM on July 19, a package containing 29 103Pd seeds with an activity of 1.54 mCi each was stolen from a delivery truck. The individual who took the container, not realizing what they were, placed the unshielded seeds under his bed. Being ill with the flu, he lay in the bed for 3 days with the seeds approximately 1 foot from his body. Calculate the dose

i d th 3 d i d

Problem 7.1


received over the 3 day period.

Neglect the attenuation of the container, the mattress and any other related items. Neglect the short periods when the individual left the bed for various reasons.

END OF


END OFMEAN LIFE

END OF


END OFCHAPTER 7

0481 - H201 - Health Physics Technology - 07 - Mean Life. · ¾Discuss the concept of effective half-life and derive it mathematically ¾Discuss significance of effective half-life

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