Question

The nuclear radius of ${}_8{O}^{16}$ is $3\times {10}^{-15}m$. If an atomic mass unit is $1.67\times {10}^{-27}kg$, then the nuclear density is approximately

• A. $2.35\times {10}^{17}gc{m}^{-3}$
• B. $2.35\times {10}^{17}kg{m}^{-3}$
• C. $2.35\times {10}^{17}g{m}^{-3}$
• D. $2.35\times {10}^{17}kgm{m}^{-3}$

Answer 1

The correct option is B $2.35\times {10}^{17}kg{m}^{-3}$

$Density=\frac{Mass}{Volume}$
$\rho =\frac{16amu}{\frac{4}{3}\pi R^3}$

$\rho =\frac{16\times 1.67\times {10}^{-27}}{\frac{4}{3}\pi R^3}$

$\rho =2.35\times {10}^{17}kg/m^3$