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: 1,570 Thanks: 613 Math Focus: Wibbly wobbly timeywimey 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. 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: 2,021 Thanks: 666 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
It's fine. Basically, you just make sure the numbers add up properly 