# Alpha emission decay: maximum possible delay

• moenste
In summary, the conversation discusses the use of a tube containing a radon isotope for medical purposes. The radon has an initial activity of 1.6 * 104 Bq and a half-life of 4 days. It decays by alpha emission and the tube is to be implanted in a patient for 8 days to deliver a specific dose. The conversation also addresses determining the proton and nucleon numbers of the daughter nucleus produced, the preference for alpha emitters over beta or gamma emitters for such purposes, and calculations for the decay constant and initial number of radioactive radon atoms in the tube. The maximum delay possible for the tube implantation to still deliver the prescribed dose is determined by considering the effects
moenste

## Homework Statement

A tube containing a isotope of radon, 22286Rn, is to be implanted in a patient. The radon has an initial activity of 1.6 * 104 Bq, a half-life of 4 days and it decays by alpha emission. To provide the correct dose, the tube, containing a freshly prepared sample of the isotope, is to be implated for 8 days.

(a) (i) What are the proton (atomic) number and the nucleon (mass) number of the daughter nucleus produced by the decay of the radon?
(a) (ii) State one reason why an alpha emitter is preferred to a beta or gamma emitter for such purposes?

(b) Determine: (i) the decay constant for radon in s-1, (ii) the initial number of radioactive radon atoms in the tube.

(c)
The operation to implant the tube has to be delayed. Ignoring the effects of any daughter products of the decay, determine the maximum delay possible if the patient is to receive the prescribe dose using the source.

Answers: (b) (i) 2.0 * 10-6 s-1, (ii) 8.0 * 109, (c) 40 hours.

2. The attempt at a solution
(a) (i) 22286Rn → 21884Po + 42α.
(a) (ii) No idea. Maybe α is less harmful for health?

(b) (i) λ = ln 2 / 345 600 = 2 * 10-6 s-1.
(b) (ii) dN / dt = - λ N → N = (dN / dt) / λ = 16 000 / 2 * 10-6 = 8 * 109 atoms.

(c) A = A0 e- λ t = 16 000 e- 2 * 10-6 * 691 200 = 4016 Bq. This should be the activity after 8 days. But how to determine the maximum possible delay? I though of something like 0 = 4016 e- 2 * 10-6 * t and solve for t, but doesn't look like mathematically solvable.

Any help please on bold problems (a) (ii) and (c)?

Last edited:
moenste said:
(a) (ii) State one reason why an alpha emitter is preferred to a beta or gamma emitter for such purposes?
It depends on what the "such purposes are", e.g. therapy vs. diagnosis. You have not specified that.
The patient has the implant for 8 days and then it is removed. This means that the patient needs a limited dose. What is that dose? This should give you a clue how to answer (c).

moenste
kuruman said:
It depends on what the "such purposes are", e.g. therapy vs. diagnosis. You have not specified that.
And in general what is the benefit?

kuruman said:
The patient has the implant for 8 days and then it is removed. This means that the patient needs a limited dose. What is that dose? This should give you a clue how to answer (c).
I though the tube "is to be implanted" = planned to be implanted. But then the operation to "implant the tube" is delayed. So the tube will be later put into the patient.

Plus if the dose is implanted for 8 days it means that the dose should be in the patient for 8 days, no more no less.

moenste said:
I though the tube "is to be implanted" = planned to be implanted. But then the operation to "implant the tube" is delayed. So the tube will be later put into the patient.

Plus if the dose is implanted for 8 days it means that the dose should be in the patient for 8 days, no more no less.
The "dose" means the total amount of radiation particles delivered. The original plan was for the brand new tube to remain for 8 days to deliver the required dose; The 8 day timeframe was for the tube when it was new. A tube with lesser initial activity would need longer to deliver the same number of particles.

moenste
moenste said:
Plus if the dose is implanted for 8 days it means that the dose should be in the patient for 8 days, no more no less.
Don't forget that the isotope decays continuously so if you wait a while, the implant will take more than 8 days to deliver the same number of alpha particles.

moenste
gneill said:
The "dose" means the total amount of radiation particles delivered. The original plan was for the brand new tube to remain for 8 days to deliver the required dose; The 8 day timeframe was for the tube when it was new. A tube with lesser initial activity would need longer to deliver the same number of particles.
kuruman said:
Don't forget that the isotope decays continuously so if you wait a while, the implant will take more than 8 days to deliver the same number of alpha particles.
Still don't understand anything.

We have something put into a person that was designed to last 8 days. So it's obvious that it should be taken from the person right away after the 8 day period.

OR, depending on the situation, the material in the person could stay for a little longer, but we need to know the activity number when the material becomes dangerous.

So the activity will be 4016 Bq on day 8. What activity level is harmful to a person? 2000 Bq? 100 Bq?

Knowing the harmful activity level we can calculate the maximum time. But we are not given it.

Other than these two situations, I don't quite see a way.

The time required to deliver a specified dose doesn't (in this problem) involve danger or harm considerations. Provided you don't give a larger dose than is specified, the amount of time it takes isn't an issue. What is important is the total dose delivered over whatever time it takes.

Here's a hint for shortcut. The dose is equal to the total number of decays that happen while the source is in the patient. That means some amount of the initial material will have decayed in that time period. Find out how much of the material decays during the original planned treatment, either by a count of atoms or just by the fraction of the original material that decays. That will be the dose that you need to be able to match in the delayed scenario, even if the source has to remain in the patient for a (possibly much) longer time.

gneill said:
Here's a hint for shortcut. The dose is equal to the total number of decays that happen while the source is in the patient. That means some amount of the initial material will have decayed in that time period. Find out how much of the material decays during the original planned treatment, either by a count of atoms or just by the fraction of the original material that decays. That will be the dose that you need to be able to match in the delayed scenario, even if the source has to remain in the patient for a (possibly much) longer time.
A0 = 16 000 Bq and A8 = 4016 Bq. Decayed over the 8 days = 11 984 Bq. Fraction decayed 0.749.

ln 0.749 = 2 * 10-6 t
t = 144 508 s = 40.1 hrs.

I still don't understand this, even though I got the correct answer.

We have activity 16 000 Bq at day 1, and activity 4016 Bq on day 8 (75 % of the material has decayed). But what's the connection between the decay percentage and the time when the material should be taken back from the patient. Maybe in some simpler words? Why should it matter what's the percentage of the decayed material?

And also:
moenste said:
(a) (ii) State one reason why an alpha emitter is preferred to a beta or gamma emitter for such purposes?
Also didn't get this part. I read some more information on α, β and γ radiation and got an understanding that α has the least penetration potential (it only penetrates skin). So it's less harmful and therefore better for medical purposes? But again, I'm pretty sure there are different procedures that require different type of radiation penetration.

moenste said:
A0 = 16 000 Bq and A8 = 4016 Bq. Decayed over the 8 days = 11 984 Bq. Fraction decayed 0.749.

ln 0.749 = 2 * 10-6 t
t = 144 508 s = 40.1 hrs.

I still don't understand this, even though I got the correct answer.
Let's take a shortcut. The half life of the material is 4 days. The initial dosing time was 8 days. That's two half-lives. So what fraction of the material "disappears" in that time? 1/2 for the first half-life, then half of what's left in the second half-life. Do if you started with N amount (atoms, mass, however you want to judge the amount), the dose was (1/2 + 1/4)N or (3/4)N. That's your 0.749 (really 0.750) fraction.

If you want to be able to deliver the same dose at a later time, and supposing you can take as much time as necessary for it all to decay, then you need to start with at least this much material. So the question becomes how long can you wait until the original material drops below 3/4 of what you started with.
We have activity 16 000 Bq at day 1, and activity 4016 Bq on day 8 (75 % of the material has decayed). But what's the connection between the decay percentage and the time when the material should be taken back from the patient. Maybe in some simpler words? Why should it matter what's the percentage of the decayed material?
As I've repeatedly stated, the dose is the total number of decays. It's not the activity itself. Lower starting activity just means it will take longer to deliver the same number of decays to the patient (dose).
And also:

Also didn't get this part. I read some more information on α, β and γ radiation and got an understanding that α has the least penetration potential (it only penetrates skin). So it's less harmful and therefore better for medical purposes? But again, I'm pretty sure there are different procedures that require different type of radiation penetration.
You're getting the picture. Usually when a source is implanted it's meant to affect a specific, local target. For example, it might be implanted next to or inside of a particular tumor. Ideally you don't want to irradiate anything other than the target. So a small penetration depth gives better control of the affected area and is a good thing when it can be used. Each radiation type has hazards for side effects according to how it interacts with the tissue molecules and atoms.

moenste

## 1. What is alpha emission decay?

Alpha emission decay is a form of radioactive decay in which an alpha particle, consisting of two protons and two neutrons, is emitted from the nucleus of an atom. This process reduces the atomic number by 2 and the mass number by 4.

## 2. How does alpha emission decay occur?

Alpha emission decay occurs when a nucleus is unstable and has too many protons or neutrons, causing it to release an alpha particle in order to become more stable. This process is random and cannot be predicted for a specific atom.

## 3. What is the maximum possible delay for alpha emission decay?

The maximum possible delay for alpha emission decay is determined by the half-life of the radioactive isotope. This is the amount of time it takes for half of the original amount of the isotope to decay. After one half-life, there is a 50% chance that an alpha particle will have been emitted. The maximum delay would occur if the atom has not decayed after multiple half-lives.

## 4. How does the half-life of an isotope affect the maximum possible delay for alpha emission decay?

The longer the half-life of an isotope, the longer the maximum possible delay would be for alpha emission decay. This is because it would take a longer amount of time for half of the original amount of the isotope to decay, increasing the chances of a longer delay.

## 5. Can the maximum possible delay for alpha emission decay be predicted?

No, the maximum possible delay for alpha emission decay cannot be predicted for a specific atom. It is a random process and can vary depending on the half-life of the isotope. However, scientists can make general predictions based on the half-life of the isotope and the amount of time that has passed since the atom was formed.

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