Binding energy og H-1

Tags:
1. Jan 14, 2016

ZeroGravity

Hello Friends !!
I have a question regarding binding energy...
Trying to calculate the binding energy of H-1 (hydrogen nucleus).
Well it is obvious that the binding energy is zero since there is no other nucleons that the proton is bound to.
But after having collected the best possible data of the atomic mass og H-1: from
http://www.physics.nist.gov/cgi-bin...lone.pl?ele=&all=all&ascii=ascii2&isotype=all
mh = 1.00782503223(9) u
one atomic mass unit:
1 u = 1.660539040 x 10-27 kg (http://physics.nist.gov/cgi-bin/cuu/Value?ukg)
The rest mass of a proton
mp = 1.672621898 x 10-27 kg (http://physics.nist.gov/cgi-bin/cuu/Value?mp)
The rest mass og an electron
me = 9.10938356 x 10-31 kg (http://physics.nist.gov/cgi-bin/cuu/Value?me)
m_nucleus = mh-me = 1.67262304224x10^-27 kg
m_defect =mp-m_nucleus =−1.14424320000x10^-33 kg

It surprises me that the result is negative and that it is so "large" a number.
I would suspect that I would get a positive number equal to the binding energy of the elevtrone ...i.e the equivalent mass og the binding energy 13.6 eV which equals E=m*c^2=> m =2.42442x10^-35 kg
This would suggest that the hydrogen atom has a mass that is 2.42442x10^-35 kg lower than the sum of the proton and electron mass which would result in a positive massdefect of 2.42442x10^-35 kg.
My result is thus a factor of about 200 "wrong"....and negative...
Any suggestions ?
Best of Greetings
Zero Gravity

2. Jan 14, 2016

Orodruin

Staff Emeritus
I suggest you double check your numerical computations.

3. Jan 14, 2016

ZeroGravity

@ Orodruin
Are you trying to tell me that I have made a mistake, or are you just guessing?
I have done the calculation a number of times, which is the reason I post here...
I do not think this is a simple matter og calculating ...
Zero Gravity

4. Jan 14, 2016

Orodruin

Staff Emeritus
Yes, you made a mistake. Correct arithmetics with your numbers give a hydrogen mass which is smaller than the mass of a proton + the mass of an electron which is much closer to the actual value

5. Jan 14, 2016

ZeroGravity

If I understand what you are saying correct...the method is ok, the numbers are ok ...
and a calculation with Ti, nSpire gives the result above...bold symbols are defined variables...

6. Jan 14, 2016

Orodruin

Staff Emeritus
What is your _amu? Dividing your original expression with the result gives 1.660540200000e-27 kg, which is not equal to the number you have quoted for 1 u.

7. Jan 14, 2016

ZeroGravity

Thanks a lot :-)
I have done this over and over again, first with the built in values of mp, me and _amu in nSpire...then looked for more accurate values...
I now get a value much closer to the expected value of 2.42*10^-35 eV which is acceptable ( for me), still curious though what causes the difference om the secont digit.

8. Jan 14, 2016

Orodruin

Staff Emeritus
Did you note the quoted errors in the value of 1 u expressed in kg? What happens when you account for it?

Also, your precision on the proton mass is just barely sufficient to get the first digit right.

9. Jan 14, 2016

ZeroGravity

I do not see errors in the atomic mass unit, but I found a value of 1 u = 1.660539040(20) x 10-27
The proton mass is quoted differently on NIST and wikipedia...even thoug wikipedia (https://en.wikipedia.org/wiki/Proton 1.672621777(74)×10−27kg) quotes NIST as a reference...
It would be nice to have a better agreement, but it seems to be a question of finding the most precise values...
the rpp-booklet2010 have entirely different values...
Where to look ?

10. Jan 14, 2016

Staff: Mentor

The "(20)" are the uncertainty on the last two digits. About 10 parts in a billion, or ~10 eV uncertainty. That is similar to the binding energy, so a deviation there should not be surprising.
On the other hand, the uncertainty on the hydrogen mass expressed in kg and on the amu expressed in kg are correlated - it is better to work with amu or eV everywhere, those measurements are more precise (and don't have those huge powers of ten).