## University Physics with Modern Physics (14th Edition)

a. Calculate the number of fissions needed. $$\frac{1.4\times10^{19}J}{(200\times10^6 eV)(1.60\times10^{-19}J/eV)}=4.38\times10^{29}$$ The mass required is $(4.38\times10^{29})(235m_p)=1.7\times10^5kg$. b. $\frac{1.7\times10^5kg }{0.007}=2.4\times10^7kg$ of mined uranium per year.