The half life for Iridium-192 is?

The half life for Iridium-192 is 74 days.

If you have a 10 Curie source of Iridium-192. In 148 days what will be the new activity of this source?

In 148 days, the activity of the Ir-192 source would be 2.5 Curies.

If you had a 10 Curie Iridium-192 source how many days would it take for the source to reach a strength of 1.25 Curies?

In 222 days the source will decay to 1.25 Curies. This is because to go from 10 curie to 1.25 curie you need to go through three half-lives. Since the half life is 74 days, simply take 74 days times 3 half-lives.

(3)(74) = 222 days.

You have a 10 Curie source of Ir-192. What will the activity be in 2,640 hours?

A 10 Curie source of Ir-192 will have the activity of 3.6 Curie in 2,640 hours.

First step in completing this problem is figuring out what equation to use. If you use the following half life equation, we can then start to put in the numbers.

A_{t}=A_{o}e^{(-.693)t/T}

A_{t} =activity at a given time

A_{o} = activity at zero time, or original activity

t = amount of time of decay

T = half life of the material in question

First thing to do is we know that Ir-192 half life value is in days. So, we either need to convert Ir-192 into half life hours or we need to convert hours into days. I prefer that latter. Simply take 2,640 hours and divide by 24 hours. This will give you days.

2,640/24 = 110 days

A_{t} = x (the unknown)

A_{o} = 10 curie

t = 110 days

T = 74 days

x = 10 curie e^{(-.693)110/74}

x = 3.6 curie

If you are having trouble getting this to work on your calculator try these steps. First, take 110 divided by 74. Then take this number and times it by -.693. Then, on you calculator find the e of this number. Most of the time you have to hit the second key on your calculator and you might be looking for a key that looks like e^{x}. Take that number and times it by 10 Curie. This is your answer!

110/74 = 1.486

(1.486)(-.693) = -1.03

-1.03 converted from the natural log = .36

(10)(.36) = 3.6 Curies

You have a 9.0 curie source of Ir-192, and the treatment time for patient A is 12 minutes. If 30 minutes is the longest acceptable treatment time allowed, then with this source how may days will it take to reach a treatment time of 30 minutes for patient A (assume nothing else changes and all other variables remain constant)

It will take 97.8 days for this source to decay to a 30 minute treatment.

To do this problem I modified the last equation. In this example we can assume time will act like activity. Thus, if a treatment time takes 12 minutes, in 74 days that same treatment should take 24 minutes. Thus we can put in the 12 minutes and 30 minutes for our activity. This will only work if the half-life value is much different then the treatment time. In this case the half life is 74 days compared to 30 minutes. If the half-life was 10 hours, this modified equation would not work. So, we can set the following up:

M_{i}=M_{f}e^{(-.693)t/T}

M_{i} = indwell time

M_{f} = maximum time allowed

t = amount of time of decay

T = half-life of the material in question

So, now we can plug the following numbers into our modified equation:

M_{i} = 12

M_{f} = 30

t = x (the unknown)

T = half-life of the material in question

12=30e^{(-.693)x/74}

If you are having trouble getting this to work on your calculator try these steps. First, take 12 divided by 30. Then take the natural log of .4. To do this on your calculator take .4 and hit the LN button. Take this number and divide by -.693. Take this number and times it by 74. This will give you how many days it takes to reach a 30 minute treatment.

.4 natural log = -.9162

-.9162/-.693 = 1.3222

(1.3222)(74) = 97.8 days

In 97.8 days, what will be the activity of this source be if it started from 9.0 Curies?

The activity of this source will be 3.6 curie in 97.8 days.

Again we can use the first equation:

A_{t}=A_{o}e^{(-.693)t/T}

A_{t} =activity at a given time

A_{o} = activity at zero time, or original activity

t = amount of time of decay

T = half life of the material in question

So, for this example we know the following:

A_{t} = x (the unknown)

A_{o} = 9 Curies

t = 97.8 days

T = 74 days

x = 9 Curies e^{(-.693)97.8/74}

If you are having trouble getting this to work on your calculator try these steps. First, take 97.8 divided by 74. Then take this number and times it by -.693. Then on you calculator find the e of this number. Most of the time you have to hit the second key on your calculator and you might be looking for a key that looks like e^{x}. Take that number and times it by 9 Curies. This is your answer!

97.8/74 = 1.321

(1.321)(-.693) = -.916

-.916 converted from the natural log = .4

(9)(.4) = 3.6 Curies

You have an unknown source of 1 Curie. In 4.3 days, this unknown source decayed to .333 Curie. What is the half-life of this unknown source

The half life of this unknown source is 2.7 days.

Again we can use the familiar equation, but in this case we will be solving for the half life value.

A_{t}=A_{o}e^{(-.693)t/T}

A_{t} =activity at a given time

A_{o} = activity at zero time, or original activity

t = amount of time of decay

T = half life of the material in question

So, for this example we know the following:

A_{t} = .333 curie

A_{o} = 1 curie

t = 4.3 days

T = x (the unknown)

.333=1e^{(-.693)4.3/x}

Simply solve for x. To set this right in your calculator, first take .333 divided by 1. This gives you a value of .333. Take this and hit the LN key on your calculator. Take this number and divided by -.693. Take this number and write it down. Take 4.3 and divided by the number you wrote down. This gives you the half life!

.333 natrual log = -1.1

-1.1/-.693 = 1.587

4.3/1.587 = 2.7 days

If the half life is 2.7 days, what do you think this unknown substance is?

This unknown substance would more than likely be Au-198 (gold).

THE END

*Author:* Adam Buell
*Co-Author:* Deb Zehel
*Editors:* Mary Hare
*Date Created:* August 2006