1
$\begingroup$

Kenneth S. Krane, Introductory to Nuclear Physics, defines mass defect as Δ=(m(A,Z)-A)$c^2$, where m is the mass of the nucleus with atomic number Z and mass number A and he says that given Δ, we can use the nuclear binding energy to deduce the atomic mass, where the nuclear binding energy is:

B=(Z$\times$mH+N$\times$mn-mA)$c^2$ , where mH is the mass of Hydrogen, mn is the mass of the neutron and mA the mass of the nucleus with atomic number Z, mass number A and N neutrons. But Wikipedia says the following:

"The mass defect of a nucleus represents the amount of mass equivalent to the binding energy of the nucleus (E=m$c^2$)"

So, what is the mass defect? The definiton from Krane's and Wikipedia's don't seem equal to me, honestly Krane's definition seems random me. Are these 2 statements equivalent? And how can the mass defect be used in the nuclear binding energy equation to deduce the atomic mass?

$\endgroup$

1 Answer 1

1
$\begingroup$

Your first formula, $\Delta=(m(A,Z)-A)c^2$ is the binding energy, which is equal to $\mu c^2$, $\mu$ being the mass defect. Here, $m(A,Z)$ is the observed mass of the nucleus, while $Zm_H+Nm_n\neq Zm+(A-Z)m=Am$ is the sum of masses of the nucleons making up the nucleus (and we assume that the proton and the neutron have equal mass $m$. For a stable nucleus, the above definition of $\Delta$ will be a negative quantity.

Yes, the two definitions are equivalent. The mass of a nucleus can be broken down into the two components: sum of masses of the constituent protons and neutrons, and the mass contributed by the binding energy ($\mu=\Delta/c^2$). $$m_A=(Zm_H+Nm_n)-\mu$$ The latter arises from the interaction between the protons and neutrons. The mass defect is precisely this second component of the mass.

If a nucleus (of mass $m_A$) were somehow dismantled completely into individual protons and neutrons (of mass $m_H, m_n$), and we assumed (as in the case of breaking a large cube of wood into smaller wooden cubes) that the sum of masses of the constituents should be equal to the mass of the whole, we would find that our assumption was off, by an amount equal to the mass defect. From this, we conclude that the interaction that holds the protons and neutrons together (binding energy $\Delta$) has to be included in the mass of the nucleus, making up the excess. This is Wikipedia's definition, $\mu=\Delta/c^2$.

$\endgroup$
6
  • $\begingroup$ But A, is the atomic mass number (N+Z), why are you saying, in your definition of A, that it is the "sum of masses of the constituent protons and neutrons, and the mass contributed by the binding energy"? I mean, if A was that, then the equation for Δ would make sense, but A is the atomic mass number... $\endgroup$
    – orochi
    Mar 26, 2020 at 0:48
  • $\begingroup$ I should have put a disclaimer: working in appropriate units. The notation was sloppy, I should have used $m_A$ for the mass. The original equation in the question uses $A$, and I wanted to be consistent with that. $\endgroup$
    – NewUser
    Mar 26, 2020 at 0:54
  • $\begingroup$ when you say, the original equation, you mean the equation for Δ? Do you mean that A in that equation is equal to the sum of the masses of the nucleus parts, and not equal to the atomic mass number? Because i still don't understand how the equation for Δ is the same as the binding energy, and i understand the equation for the binding energy, B, that i wrote in my question which is what you describe in your last paragraph. $\endgroup$
    – orochi
    Mar 26, 2020 at 1:54
  • $\begingroup$ I think you are referring to Equation 3.25 in Krane, and the section that follows. He just says that to apply 3.25, we can either use a table that gives us the mass of a nucleus, or the mass defect. Your first equation for $\Delta$ then assumes the proton and the neutron to have equal mass (say $m$), and rewrites $Zm_p+Nm_n$ as $Zm+(A-Z)m=Am$. $\endgroup$
    – NewUser
    Mar 26, 2020 at 2:24
  • $\begingroup$ I noticed I had a typo in my answer, and to clarify, $m(A,Z)$ is the observed mass of the nucleus, not the sum of nucleon masses. Also, Krane's $A$ has units of mass, it is not just a number. $\endgroup$
    – NewUser
    Mar 26, 2020 at 2:27

Your Answer

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

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