Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Azureilai on December 07, 2012, 12:46:02 PM
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So, my teacher taught us about radioactive decay in class and gave us some examples. But I have still no conception of how the formulas work. So, can someone please explain to me about Beta, Positron, Electron capture, and Alpha decay. I understand that it has to do with losing/gaining protons, neutrons, and electons, but I don't get how to write it in a formula.
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We can talk you through it, but you have to do all the work!
Let's pick a random isotope, say 53I128, since it can decay by most of those processes. a) What does the formula for the isotope mean? b) What does that tell you about the makeup of the nucleus of the atom? c) How many protons and neutrons are present?
1) Now, which decay would you like to start with? Pick one
2) Describe the process that is carried out to emit that particle. a) How would that affect the structure of the nucleus? b) What is the new number of protons and neutrons? c) What is the formula for the new isotope that is created?
3) Write all of the above into a chemical formula
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Note:
[tеx]^{128}_{\phantom{0}53}I[/tеx]
will be rendered as
[tex]^{128}_{\phantom{0}53}I[/tex]
(\phantom{0} is needed to nicely align the numbers).
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Ok, I will go with Alpha Decay.
I know it has 53 protons (since it the atomic number) 53 electrons and 128 as mass.
I know that in Alpha decay, there is the 2He4+ that is emitted, but I don't know how to write the equation.
(the 4+ represent mass. Im not sure how to render it to the front)
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Wow, that's really cool Borek! Not sure I'll be able to duplicate it though...
Okay Azureilai, your alpha particle isn't quite right. The charge should be 2+, as in [tex]^{4}_{2}He^{2+}[/tex]
Can you explain why? How many protons, electrons, and neutrons does it have?
Now if you take your original atom and remove that many protons, neutrons, and electrons, what do you have left?
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The helium atom has 4 for mass, so 2 protons and 2 neutrons, but im not sure whether it has four electrons or nonre.
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It will have no electrons, because it is being kicked out of a nucleus which has no electrons. The only way you get electrons inside the nucleus is if you have electron emission or electron capture. (And the choices would be 2 electrons or none, not 4 electrons).
But anyway, so you have your iodine nucleus which has 53 protons and a mass of 128 (how many neutrons is that?) and you remove 2 protons and 2 neutrons as an alpha particle. What do you have left?
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It has 75 neutrons, and after I remove 2 protons an neutrons, it has 124 as mass and 52 protons and 73 neutrons, and no electrons. Is that right?
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You started with 53 protons and took away 2 - that isn't 52...
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ah, sorry, meant 51
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:D Almost there - what isotope do you end up with? You already know the number of protons and the atomic mass...
And with that information you can write the reaction formula you are looking for.
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So its Antimony with no electrons and 73 neutrons.
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No electrons in the nucleus, anyway ;)
So how do you write the reaction? Starting material on the left, products on the right...
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So basically the original parent nucleus Iodine on the left, then arrow to the right, and an helium plus Antimony atom. Protons forboth of them will be put at the bottom left,mass will be at the top, so,
Iodine -- > Antimony+Helium (Except it will be written short handed with all the numbers in the right place.)
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Exactly! Now do the same thing with the other forms of nuclear reaction you asked about, starting with the same iodine parent nucleus