Level: Introductory, Intermediate

(Qu.) How do I add a new question?

(Ans.) Just click on the discussion tab then on the "+" button above and enter your question

(Qu.) How do I reply to a question?

(Ans.) Just click on the "edit" link above the question you want to answer and add you text

## How is Mo99 produced?

(Qu.) In nuclear medicine, the nuclide technetium-99m, Tc99m, is often mentioned in connection with in-vivo gamma irradiation. The parent nuclide of Tc99m is Mo99 which is produced from the thermal fission of U235. The fission yield of Mo99 is around 6%.

I would like to know the particular reaction path responsible for the production of Mo99, i.e.

$n + {}_{92}^{235}\!U = {}_{92}^{236}\!U = {}_{42}^{99}\!Mo + ? + ?n$

(Ans.) Answer from the Nucleonica team

You are interested in the production Mo99, so start by considering this as one of the two fission products. Typically the number of neutrons produced in the fission process is small, either 2 or 3. Let's assume 3, then, in order to conserve charge and mass, the reaction must be

$n + {}_{92}^{235}\!U = {}_{92}^{236}\!U = {}_{42}^{99}\!Mo + {}_{50}^{134}\!Sn + 3n$

If the number of neutron were only 2, then instead of Sn134, Sn135 would be produced.

But how important is the above reaction?

It can be seen from the fission yield module in Nucleonica (see below) that although the cumulative fission yield is 6.13%, the independent fission yields is very small – 1.80e-3%. This indicates that the above reaction path is not the main source of Mo-99 when U-235 is fissioned with thermal neutrons.

Yelds of Mo99 from the thermal fission of U235

Which other reactions give rise to Mo99?

From the nuclide chart (see figure), it can be seen from the isobar 99, that the following nuclides also give rise to Mo-99 through radioactive decay:

41Nb99, 40Zr99, 39Y99, 38Sr99, 37Rb99, 36Kr99

Exerpt from the Karlsruhe Nuclide Chart for isobar 99. All nuclides shown decay by ß- emisson (from bottom right Kr99 to the top left Tc99

From the fission yield module in Nucleonica, the independent and cumulative fission yields are shown in the figure below:

Fission yield module showing results for isobar 99. The individual and cumulative yields highlight the main nuclides resulting from U-235 fission

Here it can be seen from the independent fission yields, only the direct production of the nuclides 40Zr99 (3.99%) and 39Y99 (1.77%) is important with a smaller contribution from 41Nb99 and 38Sr99. Together, the nuclides 40Zr99 and 39Y99 give a cumulative yield of 3.99 + 1.77 = 5.76% which is close to the full value of 6.13%. The "partner" nuclides to Zr99 and Y99 are (in the case of 3 neutrons being emitted) are 52Te134 and 53I134.

Main fission reactions giving rise to Mo99

Hence the main reactions giving rise to Mo99 are:

$n + {}_{92}^{235}\!U = {}_{92}^{236}\!U = {}_{40}^{99}\!Zr + {}_{52}^{134}\!Te + 3n$

and

$n + {}_{92}^{235}\!U = {}_{92}^{236}\!U = {}_{39}^{99}\!Y + {}_{53}^{134}\!I + 3n$

The Zr99 and Y99 then decay by beta emission to Mo99.

## "Plus de Bq que d'atomes !!!!!!"

(Qu.) Monsieur, Voici les données qui comparaissent pour le Be 8 sur le site de Nucleonica. (Et il est ainsi pour les éléments de demi-vie < à la seconde).

Activité sécifique du Be 8. T 1/2: 7E-8 nanosecondes; 7,45E38 Bq/gr.

Petite vérification sur l'exactitude de l'activité spécifique proposée.

Quantité d'atomes dans 1 gr de Be 8 6,0221415E+23/8 = 7,53E22.

Conclusion: Plus de Bq que d'atomes !!!!!!

Paolo Scampa, Italy

(Ans.) Answer from the Nucleonica team

The main point in your question is that "one cannot have more Bq than atoms". In the specific example you quote that 1g Be8 has a specific activity of 7.45E38 Bq/g and that 1 g of Be8 has 7.52E22 atoms. You then continue that this is not possible since one cannot have "more Bq than atoms"

The mistake you have made is the result of a common misunderstanding. First of all, Bq is a disintegration rate (i.e. number of disintegrations per second). The number of atoms is a pure number. These two quantities cannot be compared.

The activity A(t) at any moment of time is defined by

$A(t) = \frac{dN(t)}{dt}$

where $N(t)$ is the number of atoms at time t, and $N(t) = N_0.exp(- {\lambda} t)$. At time t=0, the instantaneous activity $A_0$ is given by

$A(0) = A_0 = \frac{dN(t)}{dt}\mid _{t=0}= -\lambda} N_0$

The number of disintegrations in a given time interval needs to be calculated by integration. The activity as a function of time is given by:

$A(t) = A_0 \cdot exp(- {\lambda} t)$

where $A_0$ is the initial activity (instantaneous activity at $t=0$). To find the total number of disintegrations over a time long compared with the half-life, we have total number of disintegrations D:

$D =\int\limits_{0}^{\infty} A(t)dt = \int\limits_{0}^{\infty} A_0 \cdot e^{-\lambda t}}\,dt = N_0$

So the total number of disintegrations over a time long compared with the half-life is just equal to the total number of atoms $N_0$ as expected. Nevertheless, the activity used in this equation $A_0(t)$ is the activity at time $t=0$. In the case of the Be8 decay, this would be $\simeq 10^{38}$ Bq (for one gram of Be8).

In conclusion,

• The specific activity of Be8 is indeed 7.45E38 Bq (as shown in the figure below). This gives the instantaneous activity of 1 g of Be8.

• To find the total number of disintegrations in 1 second, one has to integrate this specific activity over time i.e.

$D =\int\limits_{0}^{\infty} A(t)dt = \int\limits_{0}^{\infty} A_0 \cdot e^{-\lambda t}}\,dt = N_0$

This equation shows clearly the relationship between the specific activity at $t=0$ i.e. $A_0$ and the total number of atoms $N_0$. Clearly the total number of disintegrations cannot exceed $N_0$ the total number of atoms. But of course the disintegrate rate can have any value (depending on the half-life).

• The results given in Nucleonica's Mass Activity calculator (shown in the figure below) are correct - and Nucleonica remains as you say "le fleuron de la physique nucléaire d'Europe".

Nucleonica's Mass Activity Calculator. Results shown are for 1 g Be8

Nucleonica's Mass Activity Calculator: Results for Be8

## Multiple decay modes on the Karlsruhe Nuclide Chart

Ir-167 from the Karlsruhe Nuclide Chart, 7th Edition, 2006

(Qu.) In the Karlsruhe Nuclide Chart, the main and subsidiary modes of decay are usually indicated by large and small triangles respectively inside the box (as in the 30.0 ms metastable state Ir-167m). However this is not the case with the nuclide Ir167 ground sate. What is the reason for this?

(Ans.) For most nuclides, there is usually one or two main decay mode (defined as modes with a branching ratio ≥ 5%) with subsidiary modes (branching ratio < 5%). In the case of Ir167 there are three decay modes with branching ratios ≥ 5%. In this case the box structure is modified to show all three modes.

The metastable state Ir-167m (30.0 ms) has the following decay modes and branching ratios:

α=0.8, ß+=0.196, p=0.004

hence the two main modes are α and ß+ (large yellow and red triangles) with the subsidiary mode p (small brown triangle, top left corner).

The ground state Ir-167 (35.2 ms) has the following decay modes and branching ratios:

α=0.48, p=0.32, ß+=0.2

since the brachnig ratios are all ≥ 5%, there are three main modes. In such a case, the decay modes are drawn as rectangles as shown.

Ir-167 from Nucleonica, 2007

In Nucleonica, the nuclide box structure is somewhat different from that of the Karlsruhe Nuclide chart. Here the box has a main mode (indicated here by the colour yellow) and subsiduary modes (indicated by the small red and brown triangles)

## Is U-234 primordial?

from Dr Jesse Shore, Senior curator, Sciences, Powerhouse Museum, Australia

Uranium-234 from the Karlsruhe Nuclide Chart, 7th Edition, 2006

(Qu.) I have been studying the 7th edition of the chart of the nuclides to assist me in developing an exhibition about nuclear science. The chart indicates uranium 234 as primordial. I understand that it is a decay product in the U238 decay series and that it occurs in nature in small amounts. Given its half-life of approximately 2x10e5 years I would have thought that none of it is primordial and instead its abundance is solely from radiogenic parents. Have I misinterpreted the chart regarding this nuclide?

Nucleonica Team:

(Ans.) That's a very interesting question. You are absolutely correct. The U-234 is not primordial. However for all practical purposes it can be treated as being such.

Due to the relatively short half-life (2.5E5 y), any U-234 present following the supernova explosion (which led to the formation of our solar system) has of course decayed a long time ago. The U-234 in uranium is produced by the radioactive decay of U-238. Since the U-234 also decays with a half.life of 2.5E5 years, the U-234 is in "radioactive equilibriium" (where the production rate is balance by the decay rate) with its parent U-238. Due to this radioactive equilibrium, the amount of U-234 in uranium is significant - approximately 0.0054%.

The black rectangle in the nuclide chart allows us to allocate an abundance to this isotope (0.0054%). Uranium obtained from an ore will contain this amount of U-234. So as I said above it is convenient to treat it as primordial.

This raises other aspects too. Why for example is Th234 not "primordial". Although it has a half-life of only 24 days, it is in radioactive equilbrium with U-238. Well the answer is just that the amount of Th-234 is neglibible. For this reason there is no black rectangle.

(Qu.) What is the effective dose obtained by inhalation of 1 µg Po210?

(Ans.) 700 Sv / µg.

## Inhalation radiotoxicity of Po-210 vs. Pu-238

(Qu.) What is the difference in inhalation radiotoxicity between Pu-238 and Po-210?

Nucleonica Team:

(Ans.) This information can be obtained form the Datasheets/ Derived Data: The inhalation radiotoxicity for Pu238 is 70 Sv /µg - a factor of only 10 lower than that for Po-210 (i.e. 700 Sv/µg).

## Coloured text in the NucleonicaWiki?

(Qu.) How do I use colours in the NucleonicaWiki?

(Ans.) This text shows the use of colours! For more information see Editing Help, HTML in Wikitext.

## What are Auger Electrons?

A good explanation of this is given in the following animation on the Auger effect

## Decay chains and branching ratios of U-234

(Qu.) Using both Nucleonica.net and directly JEF 2.2, I’m surprised to find differences in the daughters given for the 234U: Rn218, At218, Tl210, Tl206, Hg 206 are not considered in Nucleonica.net, as well as in some other references. Why? Is it linked with their half-life?

K. Beaugelin-Seiller, Institut de Radioprotection et de Sûreté Nucléaire (IRSN), France

(Ans.) Karin, these isotopes are indeed taken in account in the decay chains of U234 in Nucleonica.net. You can see this in the Nuclide Explorer by selecting U234 and with the right mouse button selecting "show decay chain". If you used the decay module to compute the chains, I would like to draw your attention to the parameter Min.Prod. (Default 1e-2) which is the minimum branching considered for the calculation. You need to select a much smaller value to obtain the nuclides you are looking for (i.e. nuclides with small branching ratios).

To demonstrate this clearly, consider the decay of Pb210 for which (from the Datasheets)

Basic decay data of Pb-210

If your Min.Prod has a value of 1e-2 the alpha decay mode will be ignored and only the Bi210 would be considered as a decay product for computing. To "see" the alpha decay mode you have to set Min.Prod. to 1e-8. If you are unsure which value to use, just insert 0 and all decay chains will be shown.

In some cases there are many decay modes, and to save calculation time, one selects a value of Min.Prod. which typically gives the most important 3-4 chains. With a default value of Min.Prod. = 1E-2, we see all chains for which the product of the branching ratios is greater than 1E-2 or 1%.

## Gamma emission branching ratio for Bi-212

(Qu.) I am involved in the gamma spectrometry measurements above ground areas with increased U-Ra- and Th-Chain content. These measurements have been performed by a number of institutions in the context of a comparison at WISMUT sites, which was organized by the BfS. In particular, I am interested in the detection of Bi-212. When I compared the branching ratios (for the emitted gamma lines) with Nucleonica.net (see table below) I found large differences.Could you please explain me why there is such a difference and where does it come from?

Basic decay data

Bernd Horlbeck, Deutsche Gesellschaft zum Bau und Betrieb von Endlagern für Abfallstoffe mbH(DBE), Germany

(Ans.) The values given in Nucleonica.net are from the JEF 2.2 datafile. The values have recently been re-assessed and in this most recent evaluation (see http://www.nucleide.org/DDEP_WG/Nuclides/Bi-212_com.pdf), the values are very close to the JEF.2.2/Nucleonica.net values. The other values you quote (RadDecay, Erdtmann, etc.) are old and no longer accurate (for more information see http://www.nucleide.org/DDEP_WG/DDEPdata.htm). Note also that the FZK evaluation that you mention is the one which is used in the Gammavision isotope standard library (and probably also for the Genie2000 standard library). This error, however, is only in the standard libraries - not in the PTB or Berkeley libraries.

PS: In the future edition of Nuclides.net, there will be a special module to create custom libraries from our databases to be used with both Genie2000 and Gammavision.

## Effective dose coefficient for Po-210 in ICRP 68 and 72

(Qu.) Po-210 is of broad practical interest: in tabacco, in drinking water, in diet, and as part of the uranium decay chain. According to the ICRP Publication 72, the effective dose coefficient for ingestion for members of the public ( eing (50)=1.2x10-6 Sv/Bq), is much higher than the value given in ICRP 68 for workers ( eing (50)=2.4x10-7 Sv/Bq). This seems to be a result of the gut transfer factor (f1) for members of the public being raised from 0.1 to 0.5. How can this be explained?

Helmut Kowalewsky, Germany

Dear Mr. Kowalewsky, you are right in saying that the gut uptake fraction (f1 value) is 5 times higher for members of the public than for workers. This is because the public can be expected to encounter Po-210 in foodstuffs rather than in the inorganic forms which are likely to occur in the workplace. ICRP Publication 67 summarises the data from ingestion of Po in reindeer and crabmeat (by humans) and in milk and liver meat (by rats) to justify the value of 0.5 chosen for the public. Of course, this leads to a higher dose coefficient (Sv/Bq) for the public than for workers, by about a factor of 5. ICRP is currently reviewing f1 values for workers. I do not know what value they are likely to choose for workers, but it remains the case that workers are exposed to different (generally inorganic) forms of Po than are the public. Alan Phipps. National Radiological Protection Board , UK Note added by the Nucleonica.net team: this higher value for the effective dose coefficient for ingestion for Po-210 has a direct consequence on the reference levels for uranium (where Po-210 is an equilibrium daughter) in radiotoxicity calculations for spent nuclear fuel. More information can be found in a recent recent publication

## Decay Heat Calculations with Nucleonica

(Qu.) We are trying to calculate the heat generation in a transport package. The package contains several nuclides. First we calculated the heat generated by each nuclide using Nucleonica.net. We used the Q-value and specific activity given in Nucleonica.net for this purpose. But then we compared these calculated values with the isotopic powers given in Nucleonica.net. The isotopic power is always lower than the heat generation we calculated – see the table below. Please, can you tell us the difference between our calculated heat generation and the given 'isotopic power' of a nuclide?

Decay heat

Ralf Steiner and Lars Niemann, Hauptabteilung Dekontaminationsbetriebe, Forschungszentrum

(Ans.) Answer from the Nucleonica.net team

1. Case of H-3:

From Nucleonica.net Datasheets: H-3 is a pure ß- emitter. The energy of the emitted ß- particle is 18.571 keV with an emission probability of 1. This agrees roughly with the Q-value you quote of 18.6 keV. Note, however, that the mean decay energies (which are used for the isotopic power calculation) are “electron” = 5.71 keV. This is approximately 1/3 of the Q-value. At a first glance, this seems to be a contradiction. It is easy to explain however when you note that in beta- emission, the following reaction occurs in the nucleus

$n \rightarrow p + \beta^- + \overline{\nu}$

i.e. a neutron is converted to a proton, a ß- particle and an antineutrino v. Note that beta emission differs from alpha emission in that the beta particle has a continuous spectrum of energies between 0 and some maximum value called the end-point energy. Also that the beta energy of 18.571 keV quoted above is the end-point energy. The average energy of the beta particle is approximately 1/3 the value of the endpoint energy. This explains the factor 1/3 observed above. Note here that approximately 2/3 of the energy i.e. 12 keV is carried away by antineutrinos.

2. Case of Fe-55:

From the Nucleonica Datasheets: Fe-55 decays by electron capture (ec). The Q-value is 231 keV. The decay reaction can be written:

$Fe55 \rightarrow Mn55 + \nu$ or $(e^- + p \rightarrow n + \nu)$

In this reaction, an inner shell electrons in Fe-55 is captured by the nucleus and combines with a proton in the nucleus to form a neutron and a neutrino. In contrast to ß- and ß+ emission, the neutrino is mono-energetic. The process of electron capture leaves a vacancy in an electron shell that is then filled immediately by electrons from higher levels cascading down. This process is characterized by the emission of x-rays. Hence the only way that the electron capture process can produce energy emission is through the x-rays or Auger electrons associated with them. The remaining energy goes into the neutrino! Hence for Fe-55, we can see from the datasheets that the average x-ray and electrons energies are 1.67 and 4.22 eV respectively. The energy difference between the Q-vale of 231 keV and the x-rays + electrons of 5.89 keV is emitted as mono-energetic neutrinos (approx. 225 keV). Hence it is incorrect to use the Q-value for the calculation of decay heat. In this particular case of electron capture in Fe55, most of the energy is emitted as an antineutrino. If one use the Q value, the decay heat will be overestimated by a factor Q/5.89 keV = 39.2. This is exactly the number you given in you table below.

3. Case of Pu-240:

In contrast to ß-, ß+, and ec, in alpha emission there are no neutrinos involved. For this reason, for decay heat calculations, the Q-value can be used.

## Se-79 half-life

(Qu.) I get some problems with Se-79 halflife. I found that in some web pages it is defined as 6.5E+04y, 1.13E+06y, and in Nucleonica it is indicated equal to 6.5E+05. What is the reason for such differences and which number is real halflife of Se-79?

Asta Brazauskaite, Lithuanian Energy Institute

(Ans.) Answer from the Nucleonica team

The measurement of the half-life of Se-79 is problematic because of its long half-life and its decay without gamma ray emission. Additionally the maximum beta- energy is only 151 keV. Many attempts have been made to determine an accurate value for this half-life, using decay counting or even particle counting. The different values you've found come from different evaluations in specific database and they might not agree depending how old the database is and how many entries the evaluator had to give an average value.The present entry in Nucleonica.net comes from the Nubase evaluation from 1997. The latest Nubase 2003 edition gives value of 295 ky. On the other hand, in the 8th Table of Isotopes, they give a value of 1.13E6 y (cut off from 1998) and a recent measurement (from 2000) gives a values of 1.24E5 y).

## Cs-137 decay gamma lines

(Qu.) I'm frequently using Nucleonica.net for shielding and dosimetry calculation, but I have one problem for calculation with radionuclide Cs-137. Starting calculation program dosimetry leads to the error "no specified ray for this nuclide". Any other radionuclides are OK. Please help me to solve this problem.

Juraj Hamza, JFM, Slovakia

(Ans.) Answer from the Nucleonica team

It comes from a common mis-interpretation of the decay of the Cs-137. I guess that you're looking for the 662 keV line. The Cs-137 parent isotope beta decays (~95%) with a 30.17y half-life to produce Ba-137m which in turn decays with a 2.55min. half-life, generating a 661.6 keV gamma ray emission.

Cs137 decay

In most decay libraries the 662 keV is then associated to the decay of Cs-137 because Ba-137m is directly produced by the decay of Cs-137 and because the half-life of Ba-137m is rather short in comparison to Cs-137, thus the two nuclides are in "equilibrium".

Therefore in your Nucleonica calculation, you should use the Ba-137m nuclide as source instead of Cs-137.

## Gamma spectrum simulator -library?

Dear colleagues,

How does the gamma spectrum simulator work? Does it use a database to simulate energy peaks, and if so, is it possible to change what library it uses? For instance i would like to compare JEFF3.1, 8th Table of Isotopes (ENSDF) etc.

Regards Jorgen Borrebaek

(Ans.) Answer from the Nucleonica team

We understand you are asking about the gamma-ray library used in the GSG, i.e. where the photon energies and emission probabilities are taken from. These are retrieved from the standard Nucleonica database using a respective function in the Nucleonica.dll. The data originates from the JEFF3.1. Currently we cannot change the photon library within the GSG. If one wants to compare libraries, it is much easier to do so using the Radiations tab in the Nucleonica's Nuclide Datasheets, i.e. to compare the raw data rather than the gamma-spectra.