- #1

Ene Dene

- 48

- 0

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.