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Q. 12.8

Calculate the binding energy per nucleon for ${ }_{10}^{20} Ne ,{ }_{26}^{56} Fe$, and ${ }_{92}^{238} U$.

Strategy We first find the binding energy of each of these nuclides using Equation (12.10) and then divide by the mass number to obtain the binding energy per nucleon.

$B\left({ }_{Z}^{A} X\right)=\left[N m_{n}+Z M\left({ }^{1} H \right)-M\left({ }_{Z}^{A} X\right)\right] c^{2}$ (12.10)

Verified Solution

\begin{aligned}B\left({ }_{10}^{20} Ne \right)=&\left[10 m_{n}+10 M\left({ }^{1} H \right)-M\left({ }_{10}^{20} Ne \right)\right] c^{2} \\=&[10(1.008665 u )+10(1.007825 u )\\&-19.992440 u ] c^{2}\left(\frac{931.5 MeV }{c^{2} \cdot u }\right) \\=& 161 MeV \\\frac{B\left({ }_{10}^{20} Ne \right)}{20 \text { nucleons }}=& 8.03 MeV / \text { nucleon } \\B\left({ }_{26}^{56} Fe \right)=&\left[30 m_{n}+26 M\left({ }^{1} H \right)-M\left({ }_{26}^{56} Fe \right)\right] c^{2}\end{aligned}

\begin{aligned}B\left({ }_{26}^{56} Fe \right)=&[30(1.008665 u )+26(1.007825 u \\&-55.934942 u ] c^{2}\left(\frac{931.5 MeV }{c^{2} \cdot u }\right) \\=& 492 MeV\end{aligned}

\begin{aligned}\frac{B\left({ }_{26}^{56} Fe \right)}{56 \text { nucleons }} &=8.79 MeV / \text { nucleon } \\B\left({ }_{92}^{238} U \right) &=\left[146 m_{n}+92 M\left({ }^{1} H \right)-M\left({ }_{92}^{238} U \right)\right] c^{2} \\&=[146(1.008665 u )+92(1.007825 u )\\&\quad-238.050783 u ] c^{2}\left(\frac{931.5 MeV }{c^{2} \cdot u }\right) \\&=1800 MeV\end{aligned}

$\frac{B\left({ }_{92}^{238} U \right)}{238 \text { nucleons }}=7.57 MeV / \text { nucleon }$

All three nuclides have a binding energy per nucleon near 8 MeV, with ${ }^{56} Fe$ having the largest binding energy per nucleon, as shown in Figure 12.7.