Some Nuclear Properties — Question 3

Question 3

The cube-root radius law implies nearly constant nuclear density

Use $R=r_0A^{1/3}$ with $r_0=1.20\,\mathrm{fm}$ and nuclear mass approximately $Am_u$, where $m_u=1.66\times10^{-27}\,\mathrm{kg}$. Treat the nucleus as a uniform sphere.

1. Derive density and evaluate it.

2. For A=125, find R and compare the volume and radius with A=8.

Question 3 - Solution

The cube-root radius law implies nearly constant nuclear density

(a) Method and calculation

$$\rho=\frac{Am_u}{(4\pi/3)r_0^3A}=\boxed{\frac{3m_u}{4\pi r_0^3}}.$$$$\boxed{\rho=2.293\times10^{17}\,\mathrm{kg/m^3}}.$$

(b) Interpretation and further calculation

$$\boxed{R_{125}=1.20(5)=6.00\,\mathrm{fm},\quad R_8=2.40\,\mathrm{fm}}.$$$$\boxed{R_{125}/R_8=2.50,\quad V_{125}/V_8=125/8=15.625}.$$

Physical check

The cancellation of A reflects nuclear saturation. Real nuclei have diffuse surfaces, and some exhibit substantial deformation.