Write a nuclear equation

  • #1

Homework Statement:

strontium 90 (Sr) is a radioactive substance, which has 52 neutrons per atom. It decays to Yttrium 90(yt) which has 51 neutrons in each atom. write a nuclear equation to show the decay of strontium 90 to Yttrium 90. Include the mass number and atomic numbers in your equation

Relevant Equations:

N/a
$$Sr\frac{90}{52}\rightarrow Yt\frac{90}{51}+e\frac{0}{-1}$$ is this correct?
 

Answers and Replies

  • #2
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No, review your lessons on how to write nucleus of atoms.
 
  • #3
$$Sr\frac{90}{52}\rightarrow \frac{0}{-1}B +Yt\frac{90}{51}$$

How about this @Gaussian97 If this is also wrong can you lead me to where I'm going wrong?
 
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  • #4
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The nuclear notations say that nuclei should be written as ## _{Z}^{A}X##.
 
  • #5
The nuclear notations say that nuclei should be written as ## _{Z}^{A}X##.
OOoooo my bad, $$Sr\frac{90}{38}\rightarrow Yt\frac{90}{39}+B\frac{0}{-1}$$

still not sure this is right but I feel it's much closer @Gaussian97
 
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  • #6
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No, apart from the format (you have to write the numbers always in the left and exactly in this way: ##_{Z}^{A}X##) you have to look at your numbers, first of all, compute the values for Z and A of all your nuclei.
 
  • #7
Z=90-52=38 A=90 For Strontium. Z=90-51=39 A=90 for Ytrtitium. The A stays the same and the Z increase indicating beta decay written as $$\frac{0}{-1}β$$ or the letter e can be used. The written equation according to this should be: $$\frac{90}{38}Sr\rightarrow \frac{90}{39}Yt+\frac{0}{-1}β$$
 
  • #8
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Well, I will suppose that when you write ##\frac{90}{38}Sr## you are trying to write ##_{38}^{90}Sr##. Then now ##Z## and ##A## are okay.
Another thing, I don't know what is the convention you use in class, but usually the ##_{Z}^{A}X## is for nuclei and atoms, not for particles alone (unless a nucleus is a particle alone, then it might be okay). But the fact is that ##\beta## is not a nucleus, so you shouldn't write it like this.
But with all that you still have two problems:
1. Check your symbols, there's one wrong.
2. As you write it, this reaction is not possible, you need something else ;)
 
  • #9
##_{38}^{90}Sr \rightarrow _{-1}^ {0}e
+ _{39}^{90}Y##


" As you write it, this reaction is not possible, you need something else ;)" not quite sure what I need to add?
 
  • #10
collinsmark
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##_{38}^{90}Sr \rightarrow _{-1}^ {0}e
+ _{39}^{90}Y##


" As you write it, this reaction is not possible, you need something else ;)" not quite sure what I need to add?
That looks right to me. :smile:

That said, technically this reaction involves an anti-neutrino. Whether or not you include the anti-neutrino in your reaction equation depends on your coursework. Often times neutrinos and anti-neutrinos are ignored, and in that case, your answer looks correct to me. [Edit: Of course, if your coursework does require acknowledging neutrinos and anti-neutrinos, then you need to reflect that in your reaction equation.]
 
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  • #11
collinsmark
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Oh, and like @Gaussian97 alludes to, sometimes the beta particle is denoted as [itex] \rm{e^-} [/itex], but that depends on your coursework.

For what it's worth, I prefer your notation of [itex] ^{\ \ 0}_{-1} \rm{e} [/itex] for nuclear reactions because it preserves balance in both the atomic numbers and the mass numbers. And if ionizations are to be considered (for whatever reason), it's easy to add the ionization count in the upper right-hand corner (e.g. [itex] ^{\ \ 0}_{-1} \rm{e}^- [/itex]) and still preserve balance there too.

Of course, you should follow the guidlines used in your coursework.
 
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  • #12
Thanks, @collinsmark, legend. Yeah my coursework doesn't need to involve neutrinos and anti-neutrinos.
 

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