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

[ussrasu]

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.

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

 

[Just some guy]

Well you seem to have the correct answer - however the mass of a proton isn't the same as the mass of a hydrogen atom, so if you have the data for the mass of a proton you'd get a more accurate answer.

 

[Astronuc]

The method is correct. Using atomic mass units includes the mass of the electrons. The binding energies of the electrons effectively cancel. There is perhaps a small error of a few eV or keV.

 

[ussrasu]

Thanks for that