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The concept of a radioactive decay chain is important in this determination. In the case of 238 U, it decays over many steps to 206 Pb. In the process, it goes through 234m Pa, 234 Pa, and 234 Th. These three isotopes have detectable gamma emissions that are capable of being used quantitatively. As can be seen in [link] , the half-life of these three emitters is much less than the half-life of 238 U. As a result, these should exist in secular equilibrium with 238 U. Given this, the ratio of activity of 238 U to each daughter products should be 1:1. They can thus be used as a surrogate for measuring 238 U decay directly via gamma spectroscopy. The total activity of the 238 U can be determined by [link] , where A is the total activity of 238 U, R is the count rate of the given daughter isotope, and B is the probability of decay via that mode. The count rate may need to be corrected for self-absorption of the sample is particularly thick. It may also need to be corrected for detector efficiency if the instrument does not have some sort of internal calibration.

Half-lives of pertinent radioisotopes in the 238 U decay chain
Isotope Half-life
238 U 4.5 x 10 9 years
234 Th 24.1 days
234m Pa 1.17 minutes

A gamma spectrum of a sample is obtained. The 63.29 keV photopeak associated with 234 Th was found to have a count rate of 5.980 kBq. What is the total activity of 238 U present in the sample?

234 Th exists in secular equilibrium with 238 U. The total activity of 234 Th must be equal to the activity of the 238 U. First, the observed activity must be converted to the total activity using Equation A=R/B. It is known that the emission probability for the 63.29 kEv gamma-ray for 234 Th is 4.84%. Therefore, the total activity of 238 U in the sample is 123.6 kBq.

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The count rate of 235 U can be observed directly with gamma spectroscopy. This can be converted, as was done in the case of 238 U above, to the total activity of 235 U present in the sample. Given that the natural abundances of 238 U and 235 U are known, the ratio of the expected activity of 238 U to 235 U can be calculated to be 21.72 : 1. If the calculated ratio of disintegration rates varies significantly from this expected value, then the sample can be determined to be depleted or enriched.

As shown above, the activity of 238 U in a sample was calculated to be 123.6 kBq. If the gamma spectrum of this sample shows a count rate 23.73 kBq at the 185.72 keV photopeak for 235 U, can this sample be considered enriched uranium? The emission probability for this photopeak is 57.2%.

As shown in the example above, the count rate can be converted to a total activity for 235 U. This yields a total activity of 41.49 kBq for 235 U. The ratio of activities of 238 U and 235 U can be calculated to be 2.979. This is lower than the expected ratio of 21.72, indicating that the 235 U content of the sample greater than the natural abundance of 235 U.

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This type of calculation is not unique to 238 U. It can be used in any circumstance where the ratio of two isotopes needs to be compared so long as the isotope itself or a daughter product it is in secular equilibrium with has a usable gamma-ray photopeak.

Determination of the age of highly-enriched uranium

Particularly in the investigation of trafficked radioactive materials, particularly fissile materials, it is of interest to determine how long it has been since the sample was enriched. This can help provide an idea of the source of the fissile material—if it was enriched for the purpose of trade or if it was from cold war era enrichment, etc.

When uranium is enriched, 235 U is concentrated in the enriched sample by removing it from natural uranium. This process will separate the uranium from its daughter products that it was in secular equilibrium with. In addition, when 235 U is concentrated in the sample, 234 U is also concentrated due to the particulars of the enrichment process. The 234 U that ends up in the enriched sample will decay through several intermediates to 214 Bi. By comparing the activities of 234 U and 214 Bi or 226 Ra, the age of the sample can be determined.

In [link] , A Bi is the activity of 214 Bi, A Ra is the activity of 226 Ra, A U is the activity of 234 U, λ Th is the decay constant for 230 Th, λ Ra is the decay constant for 226 Ra, and T is the age of the sample. This is a simplified form of a more complicated equation that holds true over all practical sample ages (on the order of years) due to the very long half-lives of the isotopes in question. The results of this can be graphically plotted as they are in [link] .

Ratio of 226 Ra/ 234 U (= 214 Bi/ 234 U) plotted versus age based on [link] . This can be used to determine how long ago a sample was enriched based on the activities of 234 U and 226 Ra or 214 Bi in the sample.

Exercise : The gamma spectrum for a sample is obtained. The count rate of the 121 keV 234 U photopeak is 4500 counts per second and the associated emission probability is 0.0342%. The count rate of the 609.3 keV 214 Bi photopeak is 5.83 counts per second and the emission probability is 46.1%. How old is the sample?

Solution : The observed count rates can be converted to the total activities for each radionuclide. Doing so yields a total activity for 234 U of 4386 kBq and a total activity for 214 Bi of 12.65 Bq. This gives a ratio of 9.614 x 10 -7 . Using [link] , as graphed this indicates that the sample must have been enriched 22.0 years prior to analysis.

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Questions & Answers

how do you translate this in Algebraic Expressions
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Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
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Damian Reply
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
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I'm interested in nanotube
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what is nano technology
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Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
good afternoon madam
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In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
anybody can imagine what will be happen after 100 years from now in nano tech world
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
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how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
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I'm interested in Nanotube
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
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Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
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the Beer law works very well for dilute solutions but fails for very high concentrations. why?
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Source:  OpenStax, Physical methods in chemistry and nano science. OpenStax CNX. May 05, 2015 Download for free at http://legacy.cnx.org/content/col10699/1.21
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