# Calculating the volume of a nucleus

1. Feb 13, 2017

### rebc

I am to calculate the volume of a nucleus, say, of Iron, with mass 55.845 amu.
I come to notice that all elements have the same value for the volume if I use the formula:
$$mass=m=m_{amu}(1.66\times10^{-27}kg/1u)$$
$$volume=V=\frac{4}{3}\pi r^3$$
$$radius=R=r_0 A^{1/3}, \ r_0=1.25\times 10^{-15}m$$
$$density=\rho = \frac{m}{V} = \frac{A(1.66\times 10^{-27})}{4/3\pi r_0^3A}=\frac{(1.66\times 10^{-27})}{4/3\pi r_0^3}=constant???$$