Question

Order of magnitude of density of uranium nucleus is
$(m_p=1.67\times 10^{-27}Kg)$

• A. $10^{20}Kg/m^3$
• B. $10^{17}Kg/m^3$
• C. $10^{14}Kg/m^3$
• D. $10^{11}Kg/m^3$

Answer 1

Answer: (B)

Radius of a nucleus is given by
$R=R_0{A}^{1/3}(Where{R}_0=1.25\times 10^{-15}m)$
$=1.25A^{1/3}\times 10^{-15}m$
Here A is the mass number and mass of the uranium nucleus
will be
$m=A{m}_p$ where $m_p=massofproton$
$=A(1.67\times 10^{-27}Kg)$
$\therefore$ Density $p=\frac{mass}{volume}=\frac{m}{\frac{4}{3}\pi R^3}$
$=\frac{A(1.67\times 10^{-27}Kg)}{A(1.25\times 10^{-15}m{)}^3}orp\approx 2.0\times 10^{17}Kg/m^3$