Question 29.5: Decaying Radium Goal Calculate the energy released during an...
Decaying Radium
Goal Calculate the energy released during an alpha decay.
Problem We showed that the { }_{88}^{226} \mathrm{Ra} nucleus undergoes alpha decay to { }_{86}^{222} \mathrm{Rn} (Eq. 29.10).
{ }_{88}^{226} \mathrm{Ra} \rightarrow{ }_{86}^{229} \mathrm{Rn}+{ }_2^4 \mathrm{He} (29.10)
Calculate the amount of energy liberated in this decay. Take the mass of { }_{88}^{226} \mathrm{Ra} to be 226.025402 \mathrm{u}, that of { }_{86}^{222} \mathrm{Rn} to be 222.017571 \mathrm{u}, and that of { }_{2}^{4} \mathrm{He} to be 4.002602 \mathrm{u}, as found in Appendix B.
Strategy This is a matter of subtracting the neutral masses of the daughter particles from the original mass of the radon nucleus.
Compute the sum of the mass of the daughter particle, m_{d}, and the mass of the alpha particle, m_{\alpha} :
m_{d}+m_{\alpha}=222.017571 \mathrm{u}+4.002602 \mathrm{u}=226.020173 \mathrm{u}
Compute the loss of mass, \Delta m, during the decay by subtracting the previous result from M_{p}, the mass of the original particle:
\begin{aligned} \Delta m & =M_{p}-\left(m_{d}+m_{\alpha}\right)=226.025402 \mathrm{u}-226.020173 \mathrm{u} \\ & =0.005229 \mathrm{u} \end{aligned}
Convert the loss of mass \Delta m to its equivalent energy in MeV:
E=(0.005229 \mathrm{u})(931.494 \mathrm{MeV} / \mathrm{u})=4.871 \mathrm{MeV}
Remark The potential barrier is typically higher than this value of the energy, but quantum tunneling permits the event to occur, anyway.