# Compute the energy of the emitted alpha for Na20 decaying to an excited state

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1. Mar 20, 2017

### llatosz

I think I got this right, I just want a second opinion to know if my concepts are correct

1. The problem statement, all variables and given/known data

20Na decays to an excited state of 20Ne through the emission of positrons of maximum kinetic energy 5.55 MeV. The excited state decays by $\alpha$ emission to the ground state of 16O. Compute the energy of the emitted $\alpha$

2. Relevant equations
$T_{\alpha}=\frac{Q}{1+\frac{m_{\alpha}}{m_{X'}}}$ where $T_{\alpha}$ is the kinetic energy of the alpha particle, $m_{\alpha}$ is the mass of that alpha particle, and $m_{X'}$ is the mass of the daughter nucleus.

$Q=\Delta mc^2$

3. The attempt at a solution
For 20Na →20Ne,
$Q_0$ = $\Delta mc^2$ = 13.8885 MeV. Of that 13.8885 MeV, 5.55 MeV goes to positrons so 13.8885-5.55 = 8.3385 MeV is left, which is the excitation energy.
For 20Ne*16O + 4He, where * denoted an excited state,
$Q_1$ = 8.3385 + $\Delta mc^2$ = 8.3385 + (-4.72987252) = 3.6086295 MeV = $Q_1$
$T_{\alpha} =\frac{Q_1}{1+\frac{m_{\alpha}}{m_{X'}}} = 2.88634 MeV = T_{\alpha}$

Now this would be the kinetic energy. Would I have to take the square root of the kinetic energy squared plus the rest mass energy squared? Or is the 2.88 my answer? I think it is the 2.88 because it should not be relativistic so rest mass energy should not matter here in this case. What are your thoughts?

2. Mar 20, 2017

### Staff: Mentor

I'm quite sure the question asks for the kinetic energy of the alpha particle, not the total energy.

3. Mar 20, 2017

### llatosz

Nah, it just says "energy"

4. Mar 20, 2017

### Staff: Mentor

Yes, and I'm quite sure that refers to the kinetic energy.

5. Mar 20, 2017

### llatosz

Oh okay. So in that case do you think I did it correctly with 2.88 MeV as my answer?

6. Mar 20, 2017

### Staff: Mentor

I didn't check the numbers, the approach is right.

7. Mar 20, 2017

### llatosz

Okay great, thank you!