My Math Forum ?>>43 (over) 20 Ca +0 (over)(-1)e?

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 January 16th, 2015, 08:16 PM #11 Senior Member   Joined: Jul 2014 From: Norway Posts: 140 Thanks: 3 Math Focus: geometry 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!
 January 17th, 2015, 02:36 PM #12 Senior Member   Joined: Jul 2014 From: Norway Posts: 140 Thanks: 3 Math Focus: geometry So would it be 43 over 19K if this is at the beginning of the problem? Sorry this stuff really confuses me still!
 January 17th, 2015, 04:34 PM #13 Math Team     Joined: May 2013 From: The Astral plane Posts: 1,518 Thanks: 580 Math Focus: Wibbly wobbly timey-wimey stuff. 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. Thanks from girlbadatmath Last edited by topsquark; January 17th, 2015 at 04:41 PM.
 January 19th, 2015, 06:07 PM #14 Senior Member   Joined: Jul 2014 From: Norway Posts: 140 Thanks: 3 Math Focus: geometry 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?
 January 28th, 2015, 02:53 AM #15 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 1,884 Thanks: 610 Math Focus: Physics, mathematical modelling, numerical and computational solutions It's fine. Basically, you just make sure the numbers add up properly Thanks from girlbadatmath

 Tags >>43, >>43, over1e