Answer
a. $1.8\times10^3 MeV$
b. $7.3\times10^2 MeV$
Work Step by Step
a. Using Figure 30-1, we note that the average binding energy per nucleon for A=238 is about 7.5 MeV. Multiply this value by the number of nucleons, 238.
$(238)(7.5 MeV)=1.8\times10^3 MeV$
b. Using Figure 30-1, we note that the average binding energy per nucleon for A=84 is about 8.7 MeV. Multiply this value by the number of nucleons, 84.
$(84)(8.7 MeV)=7.3\times10^2 MeV$
