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Kinetic energy of daughter nucleus and alpha particle from alpha decay

  • Thread starter Doscience
  • Start date
2
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1. Homework Statement
In [itex]\alpha[/itex] decay a nucleus X at rest decays to a daughter nucleus Y and an [itex]\alpha[/itex] particle. Conservation of momentum and kinetic energy gives:
[itex]M_{\alpha}v_{\alpha}+M_{Y}v_{Y}=0[/itex]

[itex]\frac{1}{2}M_{\alpha}v_{\alpha}^{2}+\frac{1}{2}M_{Y}v_{Y}^{2}=Q[/itex]

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

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

[itex]E^{\alpha}_{k}=\frac{M_{Y}}{(M_{Y}+M_{\alpha})}Q[/itex]

[itex]E^{Y}_{k}=\frac{M_{\alpha}}{(M_{Y}+M_{\alpha})}Q[/itex]


2. Homework Equations
[itex]\frac{1}{2}M_{\alpha}v_{\alpha}^{2}+\frac{1}{2}M_{Y}v_{Y}^{2}=Q[/itex]

[itex]M_{\alpha}v_{\alpha}+M_{Y}v_{Y}=0[/itex]

[itex]Q=(M(X)-M(Y)-M(\alpha))c^{2}[/itex]

[itex]E^{\alpha}_{k}=\frac{M_{Y}}{(M_{Y}+M_{\alpha})}Q[/itex]

[itex]E^{Y}_{k}=\frac{M_{\alpha}}{(M_{Y}+M_{\alpha})}Q[/itex]

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

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.
 
2
0
I managed to figure this out. I rearranged the expression for Q to get [itex]E^{\alpha}_{K}[/itex]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.
 

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