Energy Balance of Beta+ Decay of $^{22}_{11}Na$

[skrat]

Homework Statement

If we have a beta + deca: $$^{22}_{11}Na(3^+)\rightarrow ^{22}_{10}Ne(2^+)+ e^+ +\nu $$ with $t_{1/2}=2.6 years$, what is the energy balance at this decay, if you know that $^{22}_{10}Ne(2^+)$ is an excited state of $^{22}_{10}Ne$ with $1.25 MeV$ larger energy.

Homework Equations

The Attempt at a Solution

Ok, I don't know how to take into account that $^{22}_{10}Ne(2^+)$ is not ground state.

I tried like this:

$(m(Na)-m(Ne))c^2- \Delta E=-1.2MeV$ which is a complete nonsense.. Now I don't get it what am I doing wrong?

I used this: http://www.science.co.il/PTelements.asp

 

[mfb]

The right side has 1.25 MeV more energy than you would expect. Nothing magical. You would get the same result if the right side would have an additional particle with an energy of 1.25 MeV.

 

[skrat]

So...

$E=m(Na) c^2-m(Ne)c^2-m_ec^2-\Delta E$

or?

 

[mfb]

If you define $\Delta E$ to be 1.25 MeV, yes.