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)

${10}^{17}kg/m^3$

The order of magnitude of mass and volume of uranium nucleus will be
$m\simeq A(1.67\times {10}^{-27}kg)$ ($A$ is atomic number)
$V=\frac{4}{3}\pi r^3\simeq {\left[(1.25\times {10}^{-15}m)A^{1/3}\right]}^3$
Hence, $\rho =\frac{m}{v}=\frac{A(1.67\times {10}^{-27}kg)}{(8.2\times {10}^{-45}{m}^3)A}$$\simeq 2.0\times {10}^{17}kg/m^3$