Q: What is the minimum amount of energy necessary to remove a proton from the nucleus of a 42-Ca atom, thereby converting it into a 41-K atom? The former has a mass of 41.958618 u, the latter 40.961825 u, and the hydrogen atom has a mass of 1.007825 u.(adsbygoogle = window.adsbygoogle || []).push({});

My answer:

42-Ca --> 41-K + 1-H

Therefore change in mass (Delta m) = 0.011032 u

I then used this mass in E = mc^2

and got the answe E = 10.278 MeV

Can someone else try this question and see if they get the same answer? Or tell me if i have done it correctly/incorrectly

Thanks in advance

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# Minimum amount of energy necessary to remove a proton from 42-Ca atom

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