Q-value calculation - nuclear physics

[Ene Dene]

http://www.nndc.bnl.gov/masses/mass.mas03, and the missing data:
u=931.50MeV
p=1.00728u=938.28MeV
d=2.01355u=1875.63MeV
e=0.511MeV

A(a,b)B=a+A->b+B
Q=[(m(a)+m(A))-(m(b)+m(B)]MeV (1)

This can't be more simple, but in every example that I saw in my lectures different values occur. For example:
3He+3He->4He+2p (this is not 3 times He, but p+p+n), Q=12.859
Using the table of masses from a given link and (1) I get Q=13.83MeV.

The second thing I don't understand is when I have this situation:
d+p->3He+photon,
1875.63MeV+938.28MeV->2809.4MeV+photon. Is Q now equal to energy of photon?
In this case 1875.63 + 938.28 - 2809.4=4.51MeV+photon.
Btw, also in my lectures Q of this reaction is 5.493MeV.

I really can't understand these differences.

 

[Steve4Physics]

This question is 14+ years old at the time of answering. But maybe this will help someone who comes across it.

The link given in Post #1 not longer exists, so I can't check the values and calculation for the first question.

Let me address the second question.

deuteron + proton → helium-3 + gamma photon
D + p → ³He + γ

For this reaction, Q= 5.493MeV. ‘Q’ is the gamma photon energy.

(For information, this is the second reaction in the p-p chain, e.g. see https://en.wikipedia.org/wiki/Proton–proton_chain)

I suspect your calculated value of Q (4.51Mev) is wrong because you have used the mass of a ³He atom. You should have used the mass of a ³He nucleus. That would account for most of the discrepancy.

Minor edits.