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January 16th, 2015, 07:16 PM   #11
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is it still the correct answer if the 43 over 19 K goes before at the beginning of the equation? I am sorry for all the questions! This stuff is really hard for me!
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January 17th, 2015, 01:36 PM   #12
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So would it be 43 over 19K if this is at the beginning of the problem? Sorry this stuff really confuses me still!
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January 17th, 2015, 03:34 PM   #13
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Hmmm...we got a 40 going instead of a 43. This is correct:
$\displaystyle ^{43}_{20}Ca + ^0_{-1}e \rightarrow ^{43}_{19}K + \nu + \gamma$

-Dan

Edit: After consulting a chart of the nucleides I note that $\displaystyle ^{43} Ca$ is more stable than $\displaystyle ^{43} K$ so the reaction would probably be more like
$\displaystyle ^{43}_{20}Ca + ^0_{-1}e \rightarrow ^{43}_{19}K + \nu $
where the extra kinetic energy from the photon goes to the kinetic energy of the incoming electron.
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Last edited by topsquark; January 17th, 2015 at 03:41 PM.
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January 19th, 2015, 05:07 PM   #14
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Oh Sorry! yu were correct the first time I was just wondering if the outcome is changed if the 43 over 19 K ----> (trying to be an arrow) to 43 over 20Ca +0 over -1e

Basically the "answer" (43 over 19K) goes in front of this problem. Is 43 over 19 K still the best choice?
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January 28th, 2015, 01:53 AM   #15
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It's fine. Basically, you just make sure the numbers add up properly
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