Question
(III) What is the energy of the $\alpha$ particle emitted in thedecay $_{84}^{210} \mathrm{Po} \rightarrow_{82}^{206} \mathrm{Pb}+\alpha ?$ Take into account the recoil ofthe daughter nucleus.
Step 1
The momentum of the parent nucleus is zero because it is at rest. This will be equal to the sum of the momentum of the daughter nucleus (lead) and the momentum of the alpha particle. Therefore, we can write the momentum of lead is equal to the momentum of the Show more…
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(III) What is the energy of the $\alpha$ particle emitted in the decay ${ }_{84}^{210} \mathrm{Po} \rightarrow{ }_{82}^{206} \mathrm{~Pb}+\alpha ?$ Take into account the recoil of the daughter nucleus.
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