Question 25.5: Calculating the Energy of a Nuclear Reaction with the Mass-E...
Calculating the Energy of a Nuclear Reaction with the Mass-Energy Relationship
What is the energy, in joules and in megaelectronvolts, associated with the ∝ decay of _{}^{238}\text{U}?
_{92}^{238}\text{U} \longrightarrow _{90}^{234}\text{Th} + _{2}^{4}\text{He}
The nuclidic (atomic) masses in atomic mass units (u) are from Table D.5 in Appendix D:
_{92}^{238}\text{U} = 238.0508 u \quad _{90}^{234}\text{Th} = 234.0437 u \quad _{2}^{4}\text{He} = 4.0026 u
Analyze
The key concept here is the fact that, during a nuclear reaction, a loss or gain of mass is balanced by a gain or loss of energy. We need to determine the loss or gain of mass and then use the conversion factors (25.21) and (25.22) to convert the loss or gain of mass to the corresponding amount of energy.
1 \text { atomic mass unit }(u)=1.4924 \times 10^{-10} J (25.21)
1 \text { atomic mass unit (u)} = 1.4924 \times 10^{-10} J \times \frac{1 MeV }{1.6022 \times 10^{-13} J }=931.5 MeV (25.22)
The net change in mass that accompanies the decay of a single nucleus of _{}^{238}\text{U} is shown below. Note that the masses of the extranuclear electrons do not enter into the calculation of the net change in mass.
change in mass
= \text{nuclear mass of } _{90}^{234}\text{Th} + \text{nuclear mass of } _{2}^{4}\text{He} – \text{nuclear mass of } _{92}^{238}\text{U}
= [234.0437 u – (90 \times \text{mass } e^-)] + [4.0026 u – (2 \times \text{mass } e^-)] – [238.0508 u – (92 \times \text{mass } e^-)]
= 234.0437 u + 4.0026 u – 238.0508 u – 92 \times \text{mass } e^- + 92 \times \text{mass } e^-
= -0.0045 u
We can use this loss of mass and conversion factors (25.21) and (25.22) to write
E=-0.0045 u \times \frac{1.49 \times 10^{-10} J }{ u } = -6.7 \times 10^{-13} J
or
E=-0.0045 u \times \left(\frac{931.5 \text{ MeV} }{ u } \right) = -4.2 \text{ MeV}
Assess
The negative sign denotes that energy is lost in the nuclear reaction. This is the kinetic energy of the departing ∝ particle. Note that you can use either nuclear or nuclidic (atomic) masses in calculations based on a nuclear equation. The change in mass will be the same in either case because the masses of the electrons will cancel out.