Question

The formula $\mathrm{R}=(1.310–15){\mathrm{A}}^{1/3}\mathrm{m}$ can be used to calculate the radius $\mathrm{R}$ of a nucleus with mass number $\mathrm{A}$. As a result, a nucleus' mass density is on the order of: $({\mathrm{M}}_{\mathrm{prot}}={\mathrm{M}}_{\mathrm{neut}}=1.67\times {10}^{-27}\mathrm{kg})$

• A. ${10}^{17}\mathrm{kg}{\mathrm{m}}^{-3}$
• B. ${10}^{10}\mathrm{kg}{\mathrm{m}}^{-3}$
• C. ${10}^{24}\mathrm{kg}{\mathrm{m}}^{-3}$
• D. ${10}^3\mathrm{kg}{\mathrm{m}}^{-3}$

Answer 1

The correct option is A

${10}^{17}\mathrm{kg}{\mathrm{m}}^{-3}$

The explanation for the correct option:

A: ${10}^{17}\mathrm{kg}{\mathrm{m}}^{-3}$

Nuclear density:

• Nuclear density is the density of an atom's nucleus, which is approximately $2.31017\mathrm{kg}{\mathrm{m}}^{-3}$.
• Nuclear density is also used to describe circumstances where similar high concentrations exist, such as within neutron stars.
• This value is $1.25\mathrm{fm}$, with normal variances of up to $0.2\mathrm{fm}$.

$\begin{array}{rcl}{\mathrm{\rho }}_{\mathrm{nucleus}}& =& \frac{\mathrm{mass}}{\mathrm{volume}}\\ & =& \frac{\mathrm{A}}{{\displaystyle \frac{4}{3}}{{\mathrm{\pi r}}_0}^3\mathrm{A}}\\ & =& \frac{3}{4{{\mathrm{\pi r}}_0}^3}\\ & =& 2.3\times {10}^{17}\mathrm{kg}{\mathrm{m}}^{-3}\end{array}$

Final Answer: Therefore, option A ${10}^{17}\mathrm{kg}{\mathrm{m}}^{-3}$ is the correct answer.