# Kinetic energy of daughter nucleus and alpha particle from alpha decay

#### Doscience

1. Homework Statement
In $\alpha$ decay a nucleus X at rest decays to a daughter nucleus Y and an $\alpha$ particle. Conservation of momentum and kinetic energy gives:
$M_{\alpha}v_{\alpha}+M_{Y}v_{Y}=0$

$\frac{1}{2}M_{\alpha}v_{\alpha}^{2}+\frac{1}{2}M_{Y}v_{Y}^{2}=Q$

Where the Q value is the available energy found through $Q=(M(X)-M(Y)-M(\alpha))c^{2}$

Show the kinetic energy of the two decay products are given by

$E^{\alpha}_{k}=\frac{M_{Y}}{(M_{Y}+M_{\alpha})}Q$

$E^{Y}_{k}=\frac{M_{\alpha}}{(M_{Y}+M_{\alpha})}Q$

2. Homework Equations
$\frac{1}{2}M_{\alpha}v_{\alpha}^{2}+\frac{1}{2}M_{Y}v_{Y}^{2}=Q$

$M_{\alpha}v_{\alpha}+M_{Y}v_{Y}=0$

$Q=(M(X)-M(Y)-M(\alpha))c^{2}$

$E^{\alpha}_{k}=\frac{M_{Y}}{(M_{Y}+M_{\alpha})}Q$

$E^{Y}_{k}=\frac{M_{\alpha}}{(M_{Y}+M_{\alpha})}Q$

3. The Attempt at a Solution
I have tried rearranging the energy equation to get $E^{\alpha}_{k}=Q-\frac{1}{2}M_{Y}v_{Y}^{2}$
since
$E^{\alpha}_{k}=\frac{1}{2}M_{\alpha}v_{\alpha}^{2}$

Then rearranging the conservation of momentum equation and substituting in for various variables but I can't get anything that looks like the required expressions. I know this is a fairly simple algebra exercise but I just can't figure out what to do so any advice or suggestions would be appreciated.

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#### Doscience

I managed to figure this out. I rearranged the expression for Q to get $E^{\alpha}_{K}$like in my initial attempt but then I took Q out as a common factor which was the key. It was then just a matter of doing about 3 pages of algebra on what was inside the brackets.