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

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


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

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

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

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.


1 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 $(2)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.