Half-value thickness for barium 137m

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SUMMARY

The half-value thickness for gamma rays emitted by barium 137m is determined by the energy of the gamma photons, which is 662 keV. The relationship between gamma energy and the required thickness of lead to reduce this energy by half indicates that higher gamma energy necessitates greater material thickness. The discussion highlights the importance of consulting reliable sources for accurate gamma energy values and emphasizes the need to compare these values with graphical data to understand attenuation in materials.

PREREQUISITES
  • Understanding of gamma radiation and its properties
  • Familiarity with half-value layer concepts in radiation physics
  • Knowledge of lead as a shielding material
  • Ability to interpret graphs related to energy and attenuation
NEXT STEPS
  • Research the half-value layer for various materials used in radiation shielding
  • Learn about the decay processes of barium 137m and its gamma emissions
  • Explore the relationship between photon energy and attenuation coefficients
  • Study the principles of radiation detection and measurement techniques
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Students in nuclear physics, radiation safety professionals, and anyone involved in radiation shielding design or analysis.

fleetingmoment
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Homework Statement


The half-value layer (thickness) of gamma rays in lead is dependent on the energy of the gamma photons before they enter the lead. Find the graph which shows the relationship between the energy of the gamma photons (pasted below) and the half-value layer and read their energy. In your data book (we didn't get one or even a photocopy of the page), look up the energy of gamma photons emitted by barium 137* and compare it with the energy from the graph.
Barium 137 hvl over meV copy.png

Homework Equations


upload_2017-5-6_10-33-26.png


The Attempt at a Solution


Looking at the graph, it seems to be telling me that an increase in gamma energy at the entry point of a material corresponds with an increase in material thickness needed to cut this energy by half. I'm not really sure I understand the second part of the question, though. For one thing I don't have any data for the gamma energy associated with barium 137*, though I did read (it may have been on Wikipedia) that gamma energies are typically a few hundred kilo-electronvolts. For another, assuming I did manage to find the energy, what does it mean by 'compare it with the energy from the graph'? I imagine it means that I would take the stated energy and insert it into
upload_2017-5-6_10-26-28.png

as a function I(x) and then solve for I(0) and I(n), comparing the results with several points on the graph shown. I hope I am on the right track. Could someone please tell me what the gamma energy for barium 137* is?
 
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fleetingmoment said:
Could someone please tell me what the gamma energy for barium 137* is?
How about a web search on "barium 137"?
 
Hi kuruman,
I conducted several, but got sick of pages and pages of information about half-life etc. Luckily I've just tried again and lo and behold I found a page which told me the following: 'The energies of both the beta decay of cesium-137 and the subsequent gamma decay of the excited barium 137 are 512 keV and 662 keV, respectively.', which I'm going to run with.
Thank you.
 

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