# An example on computing the density of a typical nucleus

An example from my text is as such:

Compute the density of a typical nucleus, and find the resultant mass if we could manufacture a nucleus with a radius of 1cm.

I will only provide a partial working from the example since what follows is irrelevant to my confusion.

Solution:

$$\rho = \frac{m}{v} = \frac{m_p{A}}{\frac{4}{3}\pi R^{3}}$$

I'm finding it very confusing that the mass of a typical nucleus is the product of the atomic mass, A, and the mass of the proton, $$m_p.$$

I appreciate it if someone could explain the reasoning behind this.

Edit: And yes I appreciate if the constant down-voting cease. In my observation, a very rampant culture on PSE.

• Neutrons have almost the same mass as protons and the atomic mass is (at the moment) $1/12^{th}$ of the mass of a carbon 12 nucleus. The mass of nucleons varies a little depending on which nucleus they belong to, but the difference (the binding energy) from their mass when free nucleons is small. So, to a good approximation, the atomic mass $A$ is roughly the number of nucleons, whose mass is roughly $A$ times the mass of a free proton. So the SI mass of the lot is approximately $A\,m_p$. Commented May 25, 2015 at 3:41
• @WetSavannaAnimalakaRodVance That should be an answer, not a comment.
– rob
Commented May 25, 2015 at 5:01
• Compute the density of a typical nucleus, and find the resultant mass if we could produce a nucleus with a radius of 1 cm Commented Apr 5, 2023 at 1:13

Neutrons have almost the same mass as protons and the unit of atomic mass is (at the moment: see footnote) $\frac{1}{12}^{th}$ of the mass of a carbon 12 nucleus. The mass of nucleons varies a little depending on which nucleus they belong to, but the difference (the binding energy) from their mass when free nucleons is small. So, to a good approximation, the atomic mass A is roughly the number of nucleons (which itself is called the mass number)in a nucleus, whose mass is roughly A times the mass of a free proton. So the SI mass of the lot is approximately $A\,m_p$.
Footnote: There are some moves afoot to redefine SI units so that the values of the fundamental constants are exact (in the way that we now define the meter so that the value of $c$ is an exact value of $2.99792458\times 10^{8}{\rm m\,s^{-1}}$). The redefinition would fix an exact value of the Avagadro number (which is yet to be decided: it will be $6.02214x\times 10^{23}$: there are quibbles about which value of the digit $x$ would give the closest approximation to our current unit definitions), so that the mass of a $^{12}C$ nucleus would no longer be exactly 12 atomic mass units. But since the proposed change is in the $7^{th}$ significant figure of Avagadro's number, it will of course not affect the approximations in this answer.
• Since 2019, the Avogadro constant is a defined value: $\rm6.02214076×10^{23}\,mol^{-1}$ Commented Apr 5, 2023 at 1:57