# How does scintigraphy detect gamma photons from Iodine-131 disintegration?

• duchuy
In summary: I-131 and write an equation such asI-131 --> Xe-131 + beta- + gamma + antineutrinoThis ignores the (intermediate) excited state of Xe for convenience/simplicity. Many years ago I taught physics to radiography students. The equation they used would have beenI-131 --> Xe-131 + beta- + gammaSo the level of detail depends on the context. The original question was not rigorously stated.In summary, the equation I-131 --> Xe-131 + beta- + gamma + antineutrino is an oversimplification of the decay process, as it ignores the intermediate excited state of Xe. However, in certain contexts,
duchuy
Homework Statement
Determine the emmited particle
Relevant Equations
I --> Xe* + e- + ν_
Hi,
I'm struggling to understand an answer in my MCQ.
It states that a scintigraphy would detect γ photons emitted by the iodine's nucleus, and the answer was correct.
But I don't understand how it would detect γ photons from the iodine's nucleus since it disintegrates by forming Xe*, an electron and an antineutrino. So even if it is caused by an electronic rearrangement due to the appearance of a new electron, it would emit a X photon.
To a certain extent, we could say that excited Xe nucleus would emit γ photons but I really don't understand how the iodine's nucleus could emit γ photons...
Can someone please explain to me thank you!

duchuy said:
Homework Statement:: Determine the emmited particle
Relevant Equations:: I --> Xe* + e- + ν_

Hi,
I'm struggling to understand an answer in my MCQ.
It states that a scintigraphy would detect γ photons emitted by the iodine's nucleus, and the answer was correct.
But I don't understand how it would detect γ photons from the iodine's nucleus since it disintegrates by forming Xe*, an electron and an antineutrino. So even if it is caused by an electronic rearrangement due to the appearance of a new electron, it would emit a X photon.
To a certain extent, we could say that excited Xe nucleus would emit γ photons but I really don't understand how the iodine's nucleus could emit γ photons...
Can someone please explain to me thank you!
Your equation is missing mass (and atomic) numbers. Knowing which isotopes are involved is essential. And your antineutrino symbol should have a bar over it.

However, I guess you are talking about I-131. Look-up the 'decay scheme for I-131' and you will see a mix of ##\beta## and ##\gamma## radiation is produced.

duchuy
Steve4Physics said:
Your equation is missing mass (and atomic) numbers. Knowing which isotopes are involved is essential. And your antineutrino symbol should have a bar over it.

However, I guess you are talking about I-131. Look-up the 'decay scheme for I-131' and you will see a mix of ##\beta## and ##\gamma## radiation is produced.
Yes, sorry, I am talking about I-131. But I don't understand how the gamma photon is produced and I can't seem to find an explanation for the origin of the gamma photon emitted by I-131's nucleus. I see how gamma radiation is produced by the excited Xe nucleus, but not by iodine's.

duchuy said:
Yes, sorry, I am talking about I-131. But I don't understand how the gamma photon is produced and I can't seem to find an explanation for the origin of the gamma photon emitted by I-131's nucleus. I see how gamma radiation is produced by the excited Xe nucleus, but not by iodine's.
EDITED - sorry, I re-read the question and have changed my answer...

I think the statement:
"scintigraphy would detect γ photons emitted by the iodine's nucleus"
is an oversimplification to avoid describing the excited states of Xe.

The gamma photons are emitted by the excited Xe nuclei.

Last edited:
Steve4Physics said:
EDITED - sorry, I re-read the question and have changed my answer...

I think the statement:
"scintigraphy would detect γ photons emitted by the iodine's nucleus"
is an oversimplification to avoid describing the excited states of Xe.

The gamma photons are emitted by the excited Xe nuclei.
What's happening in the diagram is actually what's troubling me. Iodine's nuclei decays into an excited xenon atom. The excited xenom atom emits gamma photons to attain a lower and stable energy state by emitting gamma photons. Therefore the nuclei that emits gamma photons is the excited xenon's, not I-131, right?
From what I understand, I-131's nuclei decays by transmutating a neutron into a neutron, electron and antineutrino to attain a lower energy state due to its neutron excess, it doesn't emit gamma photons.

duchuy said:
What's happening in the diagram is actually what's troubling me. Iodine's nuclei decays into an excited xenon atom. The excited xenom atom emits gamma photons to attain a lower and stable energy state by emitting gamma photons. Therefore the nuclei that emits gamma photons is the excited xenon's, not I-131, right?
From what I understand, I-131's nuclei decays by transmutating a neutron into a neutron, electron and antineutrino to attain a lower energy state due to its neutron excess, it doesn't emit gamma photons.
I agree. (I don't know if you read my updated post #4.)

It is quite common to attribute the gamma photon to decay of I-131 and write an equation such as
##^{131}_{53}I → ^{131}_{54}Xe + \beta^- + \gamma + \overline {\nu_e}##

This ignores the (intermediate) excited state of Xe for convenience/simplicity. Many years ago I taught physics to radiography students. The equation they used would have been
##^{131}_{53}I → ^{131}_{54}Xe + \beta^- + \gamma##

So the level of detail depends on the context. The original question was not rigorously stated.

Steve4Physics said:
I agree. (I don't know if you read my updated post #4.)

It is quite common to attribute the gamma photon to decay of I-131 and write an equation such as
##^{131}_{53}I → ^{131}_{54}Xe + \beta^- + \gamma + \overline {\nu_e}##

This ignores the (intermediate) excited state of Xe for convenience/simplicity. Many years ago I taught physics to radiography students. The equation they used would have been
##^{131}_{53}I → ^{131}_{54}Xe + \beta^- + \gamma##

So the level of detail depends on the context. The original question was not rigorously stated.
Ok I see thank you so much for your help. So in case where I'm using I-131 to detect hyperthyroidism by scintigraphy, the sensors should surround the patient completely because gamma photons are emitted in all directions right? And again, sincerely, thank you sir!

duchuy said:
Ok I see thank you so much for your help. So in case where I'm using I-131 to detect hyperthyroidism by scintigraphy, the sensors should surround the patient completely because gamma photons are emitted in all directions right? And again, sincerely, thank you sir!
There is no reason that 'the sensors should surround the patient completely'. A well designed system is sensitive enough to only require a few percent of the photons to be collected.

We had a setup specifically for thyroid measurements (hyperthyroidism, hypothyroidism and ca). It consisted of two detectors (each being a scintillator + PM). Both were pointed to the thyroid - one at the front of the patient's neck and one at the back of the neck. This had the advantage of reducing errors due to patient positioning because the combined response curve was quite 'flat' in the region between the detectors.

duchuy

## 1. What is Iodine 131 β-disintegration?

Iodine 131 β-disintegration is a type of radioactive decay where an unstable atom of iodine-131 releases a beta particle (an electron) and transforms into a more stable atom.

## 2. How is Iodine 131 β-disintegration used in medicine?

Iodine 131 β-disintegration is commonly used in nuclear medicine for diagnostic and therapeutic purposes. It is used in imaging scans to detect abnormalities in the thyroid gland and also in the treatment of hyperthyroidism and thyroid cancer.

## 3. What are the risks associated with Iodine 131 β-disintegration?

The main risk of Iodine 131 β-disintegration is exposure to radiation, which can potentially damage cells and increase the risk of cancer. However, the amount of radiation released during this type of decay is relatively low and can be controlled through proper handling and safety precautions.

## 4. How long does it take for Iodine 131 to fully disintegrate?

The half-life of Iodine 131 is approximately 8 days, meaning it takes 8 days for half of the atoms in a sample to decay. It takes approximately 10 half-lives for the majority of the iodine-131 to fully disintegrate.

## 5. Can Iodine 131 β-disintegration occur naturally?

No, Iodine 131 is not found in nature and is only produced through artificial means, such as nuclear reactors or nuclear explosions. However, small traces of Iodine 131 can be found in the environment due to past nuclear accidents or testing.

• Biology and Chemistry Homework Help
Replies
1
Views
956
• Introductory Physics Homework Help
Replies
12
Views
409
• Introductory Physics Homework Help
Replies
4
Views
1K
• High Energy, Nuclear, Particle Physics
Replies
32
Views
2K
• Quantum Physics
Replies
29
Views
2K
Replies
5
Views
2K
• Introductory Physics Homework Help
Replies
4
Views
2K
• Introductory Physics Homework Help
Replies
7
Views
2K
• High Energy, Nuclear, Particle Physics
Replies
4
Views
2K
• Special and General Relativity
Replies
9
Views
1K