Converting between mass and binding energy in semi-empirical mass formula

Question

The semi-empirical mass formula is given in terms of the binding energy as:

$$ B(Z,N) = aA - bA^{\frac{2}{3}} - s \frac{(N-Z)^{2}}{A} - d \frac{Z^2}{A^{\frac{1}{3}}} - \frac{\delta}{A^{\frac{1}{2}}}$$

where $B(Z,N)$ is the binding energy of the atom and $Z$ is the proton number and $N$ is the neutron number.

I wish to rearrange this formula in terms of the mass instead of the binging energy and to write this in terms of $Z$, assuming a constant $A$. The formula relating binding energy and mass is:

$$ M(N, Z) = N m_{n} + Z(m_p + m_e) - \frac{B(Z,N)}{c^2} $$

I am told the result of this rearrangement is supposed to be

$$ M(N, Z) = \frac{1}{c^2} (\alpha - \beta Z + \gamma Z^2) $$

where

$$ \alpha = Am_nc^2 - aA + bA^{\frac{2}{3}} + sA + \delta A^{-\frac{1}{2}} $$

$$ \beta = 4s + (m_n - m_p - m_e) c^2 $$

$$ \gamma = \frac{4s}{A} + \frac{d}{A^{\frac{1}{3}}}$$

I can understand this result except for the terms involving $s$. On the $\alpha$ term I would have expected a prefactor of $1/4$, and on the $\beta$ and $\gamma$ terms I don't see the origin of the $4$ prefactor.

Answer 1

Looking only at the terms containing $s$, the binding energy goes as $$-s\frac{(N-Z)^2}{A}=-s\frac{(A-2Z)^2}{A}\tag{1}$$ and given the relation between mass and binding energy, the mass depends on $s$ as $$\frac{1}{c^2}s\frac{(A-2Z)^2}{A}=\frac{s}{c^2}\left(A+4\frac{Z^2}{A}-4Z\right).\tag{2}$$ Now compare with what we get using $$M(N, Z) = \frac{1}{c^2} (\alpha - \beta Z + \gamma Z^2).\tag{3}$$ $\alpha$ goes as $sA$ with $s$, $\beta$ goes as $4s$ and $\gamma$ as $4s/A$; so the terms containing $s$ in $(3)$ are $$\frac{1}{c^2}\left(sA-4sZ+\frac{4s}{A}Z^2\right)=\frac{s}{c^2}\left(A+4\frac{Z^2}{A}-4Z\right),\tag{4}$$ the same as in $(2)$.