# Keeping it together

**Classical Mechanics**Level 5

The "binding energy" of a nucleus is the total energy of the bonds between the nucleons (the protons and neutrons) in the nucleus. Consider a neutral atom with \(Z\) protons and \(N\) neutrons in the nucleus, and let the total nucleon number A be given by \(A=Z+N\). The binding energy can be written as a function of \(Z\) and \(A\) as:

\(E_{(A,Z)}=a_vA-a_sA^{2/3}-a_cZ^2A^{-1/3}-\) \(a_a(A-2Z)^2A^{-1}/4+\delta A^{-1/2}\).

The values for the parameters in the above equation are \(a_v=15.67~\mbox{MeV}, a_s=17.23~\mbox{MeV}, a_c=0.714~\mbox{MeV}, a_a=93.15~\mbox{MeV}\).

\(\delta\) can take different values depending on whether A and Z are even or odd. To simplify the calculation, we'll let \(\delta = 0\), which is one of the allowed values.

The nucleus is the most stable when the binding energy is highest, as that's when it takes the most energy to rip it apart. When \(A\) is small, we get \(Z=A/2\) for the most stable nuclei, and this explains why for lighter atoms half of the nucleus is generally protons and the other half is made from neutrons. When \(A\) is large, the most stable nuclei (the ones with highest binding energy) have \(Z \sim A^{1-y} = A^w\). Find \(w\).

**Details and assumptions**

- Take the true large A limit. Real nuclei of course stop occurring in nature as A increases but if you look at the periodic table you will start to see the trend towards the value of \(w\) you calculate.

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