Question 29.6: The Beta Decay of Carbon-14 Goal Calculate the energy releas...
The Beta Decay of Carbon-14
Goal Calculate the energy released in a beta decay.
Problem Find the energy liberated in the beta decay of { }_{6}^{14} \mathrm{C} to { }_{7}^{14} \mathrm{~N}, as represented by Equation 29.13.
{ }_6^{14} \mathrm{C} \rightarrow{ }_7^{14} \mathrm{~N}+\mathrm{e}^{-} (29.13)
That equation refers to nuclei, while Appendix B gives the masses of neutral atoms. Adding six electrons to both sides of Equation 29.13 yields
{ }_{6}^{14} \mathrm{C} \text { atom } \rightarrow{ }_{7}^{14} \mathrm{~N} \text { atom }
Strategy As in preceding problems, finding the released energy involves computing the difference in mass between the resultant particle(s) and the initial particle(s) and converting to \mathrm{MeV}.
Obtain the masses of { }_{6}^{14} \mathrm{C} and { }_{7}^{14} \mathrm{~N} from Appendix B and compute the difference between them:
\Delta m=m_{\mathrm{C}}-m_{\mathrm{N}}=14.003242 \mathrm{u}-14.003074 \mathrm{u}=0.000168 \mathrm{u}
Convert the mass difference to \mathrm{MeV} :
E=(0.000168 \mathrm{u})(931.494 \mathrm{MeV} / \mathrm{u})=0.156 \mathrm{MeV}
Remarks The calculated energy is generally more than the energy observed in this process. The discrepancy led to a crisis in physics, because it appeared that energy wasn’t conserved. As discussed below, this crisis was resolved by the discovery that another particle was also produced in the reaction.