Are the Energy Units in the Decay Diagram Correct for Pt-125?

Click For Summary
The discussion centers on the accuracy of energy units in the decay diagram for Pt-125. Participants identify discrepancies in the stated energy levels, specifically between 0.0309keV and 0.1298keV, highlighting that the latter is the correct interpretation. It is noted that the model-answer incorrectly labels 0.0309keV as an energy level, when it actually represents the transition energy between two excited states. Additionally, there is a suggestion to verify the characteristic X-ray energies for Pt-125 to ensure relevance to the energy range of interest. Overall, the conversation reveals errors in the solution's presentation of energy levels and emphasizes the need for clarity in scientific communication.
Graham87
Messages
72
Reaction score
16
Homework Statement
See picture
Relevant Equations
See picture
1.png


In the solution below it says 22% goes to 0.0309keV. From the diagram above I interpret 22% goes to 0.1298keV with EC(L)/EC(K)=3.0 and not 4.4. Why is that wrong ?

Thanks alot!

2.png
 
Last edited:
Physics news on Phys.org
Graham87 said:
View attachment 322938
In the solution below it says 22% goes to 0.0309keV. From the diagram above I interpret 22% goes to 0.1298keV with EC(L)/EC(K)=3.0 and not 4.4. Why is that wrong ?
View attachment 322939
Not sure how much I can help but there are some mistakes worth (IMO) mentioning.

Two excited nuclear energy levels (of Pt-125) are given as 0.1298MeV (=129.8keV) and 0.0989MeV (=98.9keV). Note that the difference between these is 0.0309MeV (= 30.9keV).

But there are then various references (including in the model-answer) to ‘0.1298keV’, ‘0.0989keV’ and ‘0.0309keV’. It appears that units are very messed up.

Also, the model-answer refers to an ‘energy level of 0.0309keV’. Even after correcting the unit, this value is not an energy level; it is the transition energy between the two excited nuclear states. What the model answer is telling you is that for each electron-capture, there are two possible nuclear emissions: a ##\gamma## photon of energy 30.9keV or one of 98.9keV. Note that only one of these is in the energy-range of interest.

I’m not familiar with some of the symbols used (despite several years of being a medical physicist in radiotherapy a long time ago). But I would suggest that you check the energies of characteristic X-rays for Pt-125 to see which ones are in the energy range of interest.

Edit: several minor changes.
 
Last edited:
Steve4Physics said:
Not sure how much I can help but there are some mistakes worth (IMO) mentioning.

Two excited nuclear energy levels (of Pt-125) are given as 0.1298MeV (=129.8keV) and 0.0989MeV (=98.9keV). Note that the difference between these is 0.0309MeV (= 30.9keV).

But there are then various references (including in the model-answer) to ‘0.1298keV’, ‘0.0989keV’ and ‘0.0309keV’. It appears that units are very messed up.

Also, the model-answer refers to an ‘energy level of 0.0309keV’. Even after correcting the unit, this value is not an energy level; it is the transition energy between the two excited nuclear states. (What the model answer is telling you is that for each electron-capture, there are two possible nuclear emissions: a ##\gamma## photon of energy 30.9keV or one of 98.9keV. Note that only one of these is in the energy-range of interest.

I’m not familiar with some of the symbols used (despite several years of being a medical physicist in radiotherapy a long time). But I would suggest that you check the energies of characteristic X-rays for Pt-125 to see which ones are in the energy range of interest.

Edit: minor changes only.
Thanks! Yes, you are right. Aparently the the solution is wrong. They wrote the wrong numbers above and used the right numbers during the calculations lol.
Thanks alot for the explanation!
3.png
 
At first, I derived that: $$\nabla \frac 1{\mu}=-\frac 1{{\mu}^3}\left((1-\beta^2)+\frac{\dot{\vec\beta}\cdot\vec R}c\right)\vec R$$ (dot means differentiation with respect to ##t'##). I assume this result is true because it gives valid result for magnetic field. To find electric field one should also derive partial derivative of ##\vec A## with respect to ##t##. I've used chain rule, substituted ##\vec A## and used derivative of product formula. $$\frac {\partial \vec A}{\partial t}=\frac...

Similar threads