Question

Radius of nucleus may be given as $R_N=R_0.A^{1/3}$ , where $A=massnumber$ and $R_0=Constant$ . Calculate the density of nucleus of an atom if $R_0=1.2\times {10}^{-15}$
$[N_A=6times{10}^{23}\times 8\pi \times (1.2{)}^3$ = $1000$

Answer 1

$\text{Density of nucleus}=\frac{\text{mass of nucleus}}{\text{volume of nucleus}}=\frac{A}{\frac{4}{3}\pi R_N^3}=\frac{A}{\frac{4}{3}\pi {\left(R_{\circ }{A}^{\frac{1}{3}}\right)}^3}=\frac{3}{4\pi R_{\circ }^3}=\frac{3}{4\pi (1.2\times {10}^{-15}{)}^3}=1.4\times {10}^{44}{m}^{-3}$