Question 25.3: Iron and Uranium Use data in Appendix D to compute the bindi...
Iron and Uranium
Use data in Appendix D to compute the binding energy per nucleon of { }^{56} Fe \text { and }{ }^{235} U .
ORGANIZE AND PLAN From A and Z, N=A-Z. Then binding energy is computed as in the preceding example: E_{ b }=(\Delta m) c^{2}= \left( Zm _{p}+N m_{n}+Z m_{e}-M\left({ }^{A} X\right)\right) c^{2}. We’ll divide by the number of nucleons A to get the binding energy per nucleon.
\text { Known: } M\left({ }^{56} Fe \right)=55.93494 u ; M\left({ }^{235} U \right)=235.043923 u ; m_{n}= 1.00866 u ; m_{p}=1.00728 u ; m_{e}=0.00054858 u ; Z=26, A=56\left({ }^{56} Fe \right) ; Z=92, A=235\left({ }^{235} U \right) ; 1 u \cdot c^{2}=931.5 MeV.
For iron, the binding energy is
E_{ b }=((26)(1.00728 u )+(30)(1.00866 u )+(26)(0.00054858 u ) -55.93494 u ) c^{2}=0.52840 u \cdot c^{2}.
Converting to MeV per nucleon,
E_{ b } / A=\left(0.52840 u \cdot c^{2} \times \frac{931.5 MeV }{ u \cdot c^{2}}\right) /(56 \text { nucleons }).
= 8.79 MeV/nucleon.
Similar calculations for uranium give a total binding energy of 1784 MeV, or 7.59 MeV/nucleon Both per nucleon values agree with Figure 25.5.
REFLECT The corresponding mass defect is significant: For { }^{235} U it’s almost 2 u, nearly the equivalent of two missing nucleons! The mass defect is obvious in nuclei, which is why E = mc² is often mistakenly considered to be only about nuclear physics. There’s also a mass defect when, for example, hydrogen and oxygen atoms join to form a water molecule. But it’s the mass equivalent of a few electronvolts, not mega-electronvolts; thus it’s so small compared with the molecular mass as to be unnoticeable.
