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Chapter 12

Q. 12.6

Show that Equation (12.18) can be written as

\Delta E_{\text {Coul }}=\frac{3}{5} \frac{Z(Z-1) e^{2}}{4 \pi \epsilon_{0} R} (2.18)

\Delta E_{\text {Coul }}=0.72[Z(Z-1)] A^{-1 / 3} MeV (12.19)

and use this equation to calculate the total Coulomb energy of { }_{92}^{238} U.

Strategy We use Equation (12.2) for R (with r_{0}=1.2 fm) and insert it into Equation (12.18) to find the energy first in joules and then in MeV.

R=r_{0} A^{1 / 3} (12.2)

Step-by-Step

Verified Solution

\begin{aligned}\Delta E_{\text {Coul }}=& \frac{3}{5}[Z(Z-1)]\left(1.6 \times 10^{-19} C \right)^{2} \\& \times\left(9 \times 10^{9} N \cdot m ^{2} / C ^{2}\right) \frac{1}{1.2 \times 10^{-15} m \cdot A^{1 / 3}} \\=& 1.15 \times 10^{-13}[Z(Z-1)] A^{-1 / 3} J \frac{1 MeV }{1.6 \times 10^{-13} J } \\=& 0.72[Z(Z-1)] A^{-1 / 3} MeV\end{aligned}

 

Now we insert Z = 92 and A = 238 into Equation (12.19) to find

 

E_{\text {Coul }}=0.72(92)(91)(238)^{-1 / 3}=970 MeV

 

Is this large or small? If we look up the masses in Appendix 8, we calculate that the total binding energy of 2 \underset{92}{238} U with respect to dissociation into its component nucleons is

 

\begin{aligned}B_{ tot }=&[146(1.008665 u )+92(1.007825 u )\\&\quad-238.050783 u ] c^{2}\left(\frac{931.5 MeV }{c^{2} \cdot u }\right)=1802 MeV\end{aligned}

 

The total Coulomb energy is a significant fraction of the binding energy of a large nucleus.