Question 43.2: Applying the Semiempirical Binding-Energy Formula The nucleu...
Applying the Semiempirical Binding-Energy Formula
The nucleus {}^{64}Zn has a tabulated binding energy of 559.09 MeV. Use the semiempirical binding-energy formula to generate a theoretical estimate of the binding energy for this nucleus.
Conceptualize Imagine bringing the separate protons and neutrons together to form a {}^{64}Zn nucleus. The rest energy of the nucleus is smaller than the rest energy of the individual particles. The difference in rest energy is the binding energy.
Categorize From the text of the problem, we know to apply the liquid-drop model. This example is a substitution problem.
For the {}^{64}Zn nucleus, Z = 30, N = 34, and A = 64. Evaluate the four terms of the semiempirical binding-energy formula:
\begin{aligned}& C_1 A=(15.7 MeV)(64)=1 005 MeV \\& C_2 A^{2 / 3}=(17.8 MeV)(64)^{2 / 3}=285 MeV \\& C_3 \frac{Z(Z-1)}{A^{1 / 3}}=(0.71 MeV) \frac{(30)(29)}{(64)^{1 / 3}}=154 MeV \\& C_4 \frac{(N-Z)^2}{A}=(23.6 MeV) \frac{(34-30)^2}{64}=5.90 MeV\end{aligned}Substitute these values into Equation 43.3:
E_b=C_1 A-C_2 A^{2 / 3}-C_3 \frac{Z(Z-1)}{A^{1 / 3}}-C_4 \frac{(N-Z)^2}{A} (43.3)
E_b=1 005 MeV-285 MeV-154 MeV-5.90 MeV=560 MeVThis value differs from the tabulated value by less than 0.2%. Notice how the sizes of the terms decrease from the first to the fourth term. The fourth term is particularly small for this nucleus, which does not have an excessive number of neutrons.