- #1

llatosz

- 62

- 9

1. Homework Statement

1. Homework Statement

^{20}Na decays to an excited state of

^{20}Ne through the emission of positrons of maximum kinetic energy 5.55 MeV. The excited state decays by ##\alpha## emission to the ground state of

^{16}O. Compute the energy of the emitted ##\alpha##

## Homework 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##

## The Attempt at a Solution

For

^{20}Na →

^{20}Ne,

##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

^{20}Ne

^{*}→

^{16}O +

^{4}He, 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?