Calculating the volume of a nucleus

[rebc]

Homework Statement

I am to calculate the density of a nucleus, say, of Iron, with mass 55.845 amu.

Homework Equations

see below

The Attempt at a Solution

I come to notice that all elements have the same value for the volume if I use the formula:
$$mass=m=m_{amu}(1.66\times10^{-27}kg/1u)$$
$$volume=V=\frac{4}{3}\pi r^3$$
$$
radius=R=r_0 A^{1/3}, \ r_0=1.25\times 10^{-15}m
$$

$$density=\rho = \frac{m}{V} = \frac{A(1.66\times 10^{-27})}{4/3\pi r_0^3A}=\frac{(1.66\times 10^{-27})}{4/3\pi r_0^3}=constant?$$

 

[jbriggs444]

$$
radius=R=r_0 A^{1/3}, \ r_0=1.25\times 10^{-15}m
$$

As I read this, you are not measuring the radius. You are inferring the radius from the cube root of the atomic weight. It should be no surprise that when you then cube this radius you get something that is proportional to atomic weight. i.e. a constant figure for density.

What is the origin of that formula for radius?

 

[rebc]

I have just found it on wiki. How must I find the density then? how am I to find the radius of the iron nuclei to apply on the volume equation?

 

[jbriggs444]

I may have misunderstood the thrust of your question. You expressed surprise ("?") that the density that came out of your formula was a constant. If we are to trust this Wiki page then the formula that you quote is indeed derived by an assumption of a constant density and there is no need for surprise:

"The stable nucleus has approximately a constant density and therefore the nuclear radius R can be approximated by the following formula,
$R=r_{0}A^{\frac{1}{3}}$
"

You've already written down an equation for the density.

 

[rebc]

I may have misunderstood the thrust of your question. You expressed surprise ("?") that the density that came out of your formula was a constant. If we are to trust this Wiki page then the formula that you quote is indeed derived by an assumption of a constant density and there is no need for surprise:

"The stable nucleus has approximately a constant density and therefore the nuclear radius R can be approximated by the following formula,
$R=r_{0}A^{\frac{1}{3}}$
"

You've already written down an equation for the density.

Yes I get what you're saying now. Now how do I get the radius for the density calculation?

 

[jbriggs444]

Yes I get what you're saying now. Now how do I get the radius for the density calculation?

Again, I do not understand what you are asking. You have already written down a formula for density, complete with all constants filled in.

 

[rebc]

Again, I do not understand what you are asking. You have already written down a formula for density, complete with all constants filled in.

Oh sorry, I was to compare the densities of several elements(of varying masses, of course). But now using this formula for the radius, we end up with a constant which applies for any mass(as you've explained previously, the radius equation is derived). So now how am I to find the varying radii of nuclei of several elements?

 

[jbriggs444]

Oh sorry, I was to compare the densities of several elements(of varying masses, of course). But now using this formula for the radius, we end up with a constant which applies for any mass(as you've explained previously, the radius equation is derived). So now how am I to find the varying radii of nuclei of several elements?

You cannot use an approximate formula that is based on an assumption of constant density to determine the densities of various elements and expect to come up with anything but the same density for all of them.

You are going to need to either perform a measurement yourself, find a measurement made by someone else or find a formula distilled from such measurements. I do not have anything to offer.