Question 28.QE.7: Kinetic energy produced during alpha decay Determine the kin......
Kinetic energy produced during alpha decay
Determine the kinetic energy of the product nuclei when polonium-212 undergoes alpha decay. The masses of the nuclei involved in the decay are m m\left({ }_{84}^{212} \mathrm{Po}\right)=211.9889 \mathrm{u} m\left({ }_{82}^{208} \mathrm{~Pb}\right)=207.9766 \mathrm{u}, \text { and } m\left({ }_2^4 \mathrm{He}\right)=4.0026 \mathrm{u}
Represent mathematically The alpha decay of polonium- 212 is
{ }_{84}^{212} \mathrm{Po} \rightarrow{ }_{82}^{208} \mathrm{~Pb}+{ }_2^4 \mathrm{He}+\text { energy }
Use Eq. (28.5) to find the energy Q released during
this reaction:
Q=\left(\sum_{\text {reactants }} m-\sum_{\text {products }} m\right) c^2 (28.5)
Q=\left[m\left({ }_{84}^{212} \mathrm{Po}\right)-m\left({ }_{82}^{208} \mathrm{~Pb}\right)-m\left({ }_2^4 \mathrm{He}\right)\right] c^2
Solve and evaluate Insert the appropriate values and convert to MeV:
Q = (211.9889 u – 207.9766 u – 4.0026 u)c²
=(0.0097 \mathrm{u}) c^2\left(\frac{931.5 \mathrm{MeV}}{\mathrm{u} \cdot c^2}\right)
= 9.0 MeV
Of this 9.0 MeV of released energy, 8.8 MeV is converted to the kinetic energy of the alpha particle \left({ }_2^4 \mathrm{He}\right) . Most of the remaining 0.2 MeV is converted to the kinetic energy of the recoiling lead-208 nucleus. In addition, a small fraction of the energy may be released as an additional product in the reaction: a gamma ray (a high energy photon).
Try it yourself: Determine the product nucleus when thorium-232 \left(\begin{array}{c} 232 \\ 90 \end{array} \mathrm{Th}\right) undergoes alpha decay.
Answer: Radium-228 \left({ }_{88}^{228} \mathrm{Ra}\right)