Skip to main content

curated data - Find elements in a decay chain using IsotopeData


I am trying to write a program using the Mathematica function IsotopeData, but am finding the code troublesome. Due to my (relative) familiarity with Python, I am puzzled as to how Mathematica works, as they are very different. $\require{mhchem}$


What I am trying to do, is to find the elements in a decay chain which end with either $\ce{^{191}_{77}Ir}$ or $\ce{^{193}_{77}Ir}$.


In Python syntax, what I'm trying to do would look a bit like this:


#A would be the Atomic Mass and Z the Atomic Number
#DaughterNuclide(Z,A) would return the Daughter Nuclides of the isotope

def IsoCheck(z,a):
try:

x = DaughterNuclide(z,a)
if x in Isotopes:
return [z,a]
else:
return [0,0]
except:
return [0,0]

L=[]
Isotopes = [[76,191],[78,191]]

for a in xrange(1,295):
for z in xrange(1,119):
if IsoCheck(z,a) != [0,0]:
L.append(IsoCheck(z,a))

repeat = 1
#Then, repeat until all chains are over
while repeat == 1:
repeat = 0
for x in L:

if IsoCheck(x) != [0,0] #Lets just pretend it isn't a list
L.append(x)
repeat = 1

Now, my question is this: what would the equivalent code to perform this task in Mathematica look like?



Answer



It is a nice application for the Graph[] features in Mma.
We can calculate quickly all possible decays for all known isotopes, and then let VertexComponent[] look for the chains ending in {"Iridium191", "Iridium193"}.


g = Graph@Union@Flatten[Thread[DirectedEdge @@ ##] & /@ 
Select[{#, IsotopeData[#, "DaughterNuclides"]} & /@ IsotopeData[], #[[2]] != {} &]];


Union@Flatten[VertexComponent[g, #] & /@ {"Iridium191", "Iridium193"}]

$\begin{array}{l} Actinium207&Actinium209&Astatine195\\ Astatine197&Astatine199&Astatine201 \\ Bismuth191&Bismuth193&Bismuth195 \\ Bismuth197&Francium199&Francium201 \\ Francium203&Francium205&Gold191 \\ Gold193&Iridium191&Iridium193 \\ Lead191&Lead193&Mercury191 \\ Mercury193&Osmium191&Osmium193 \\ Platinum191&Platinum193&Polonium191 \\ Polonium193&Polonium195&Polonium197 \\ Protactinium213&Radium203&Radium205 \\ Radon195&Radon197&Radon199 \\ Radon201&Rhenium191&Rhenium193 \\ Thallium191&Thallium193&Thorium209 \\ \end{array}$


Edit


The possible decay chains are:


g1 = Union[Flatten[VertexComponent[g, #] & /@ #], #] &@{"Iridium191", "Iridium193"}
g2 = Subgraph[g, g1, VertexShapeFunction -> "Name", GraphLayout -> "LayeredDrawing"]

Mathematica graphics



Edit 2


Another application.


(*All possible decays of all Isotopes *)
decays = Select[{#, IsotopeData[#, "DaughterNuclides"]} & /@ IsotopeData[], #[[2]] != {} &];
(*Identify the Isotope with more ways to decay *)
mostModes = SortBy[decays, -Length@#[[2]] &][[1, 1]];
(*Get its decay characteristics*)
mMdecays = IsotopeData[mostModes, #] & /@ {"DaughterNuclides", "DecayModeSymbols", "BranchingRatios"};
(*Aux Function*)
pos[mostModes] = Above; Table[pos[i] = Below, {i, mMdecays[[1]]}];

(*Draw a scheme of its decays modes and percentages*)
g = Framed@Graph[Labeled[#, Placed[{Text@Style[#, 14, FontFamily -> "Helvetica"]},
{pos[#]}]] & /@ Join[{mostModes}, mMdecays[[1]]],
Labeled[DirectedEdge[mostModes, #[[1]]], Placed[{ToString@StandardForm@#[[2]] <> "\n" <>
ToString[100 #[[3]]] <> "%"}, {"Middle"}]] & /@ (Transpose@ mMdecays),
ImagePadding -> 30]

Mathematica graphics


Comments

Popular posts from this blog

plotting - How to draw lines between specified dots on ListPlot?

I would like to create a plot where I have unconnected dots and some connected. So far, I have figured out how to draw the dots. My code is the following: ListPlot[{{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4,13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full] I have thought using ListLinePlot command, but I don't know how to specify to the command to draw only selected lines between the dots. Do have any suggestions/hints on how to do that? Thank you. Answer One possibility would be to use Epilog with Line : ListPlot[ {{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4, 13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full, Epilog -> { Line[ ...

dynamic - How can I make a clickable ArrayPlot that returns input?

I would like to create a dynamic ArrayPlot so that the rectangles, when clicked, provide the input. Can I use ArrayPlot for this? Or is there something else I should have to use? Answer ArrayPlot is much more than just a simple array like Grid : it represents a ranged 2D dataset, and its visualization can be finetuned by options like DataReversed and DataRange . These features make it quite complicated to reproduce the same layout and order with Grid . Here I offer AnnotatedArrayPlot which comes in handy when your dataset is more than just a flat 2D array. The dynamic interface allows highlighting individual cells and possibly interacting with them. AnnotatedArrayPlot works the same way as ArrayPlot and accepts the same options plus Enabled , HighlightCoordinates , HighlightStyle and HighlightElementFunction . data = {{Missing["HasSomeMoreData"], GrayLevel[ 1], {RGBColor[0, 1, 1], RGBColor[0, 0, 1], GrayLevel[1]}, RGBColor[0, 1, 0]}, {GrayLevel[0], GrayLevel...

list manipulation - Selecting multiple columns from a matrix?

Sample data: data = { {{2013, 1, 1}, 24.13, 167.67, 231.82}, {{2013, 1, 2}, 32.15, 170.92, 225.99}, {{2013, 1, 3}, 35.43, 172.68, 221.67}, {{2013, 1, 4}, 36.73, 173.05, 218.32}, {{2013, 1, 5}, 58.19, 165.96, 197.05}, {{2013, 1, 6}, 69.99, 163.50, 187.52}, {{2013, 1, 7}, 71.37, 154.21, 175.58}, {{2013, 1, 8}, 72.51, 149.66, 163.25}}; I want a DateListPlot with three graphs, so for a matrix formed by columns 1 and 2, one for columns 1 and 3, and 1 for columns 1 and 4. At the moment I'm using this code: data2 = Transpose[{data[[All, 1]], data[[All, 2]]}]; data3 = Transpose[{data[[All, 1]], data[[All, 3]]}]; data4 = Transpose[{data[[All, 1]], data[[All, 4]]}]; DateListPlot[{data2, data3, data4}, Joined -> True, Filling -> {3 -> {1}}] but I have a hunch that this can be done more efficiently. I don't like the Transpose s in particular. Any ideas? edit (for extra credit) What if I need to multiply the second column by 2, which in my solution is simp...