Chapter 32 - Nuclear Physics and Nuclear Radiation - Problems and Conceptual Exercises - Page 1153: 50

$1.6\times 10^5$gallons

We know that $E=\frac{m_U}{M_U}E_U$ $\implies E=\frac{1.0lb}{235u}(\frac{1Kg}{2.3lb})(\frac{1u}{1.66\times 10^{-27}Kg})(173\times 10^6eV)$ $E=2.05\times 10^{23}eV(1.60\times 10^{-19}J/eV)=3.2\times 10^{13}J$ Now we can find the required number of gallons of gas $N=\frac{E}{E_{gasoline}}$ We plug in the known values to obtain: $N=\frac{3.2\times 10^3}{2.0\times 10^8}$ $N=1.6\times 10^5$gallons