Question

Calculate the density of the\(_{12}^{24}Mg\)nucleus in\(g/mL\), assuming that it has the typical nuclear diameter of \(1 \times 1{0^{ - 13}}\;cm\)and is spherical in shape.

Short answer

\(\rho = 7.66 \times {10^{16}}{\rm{g/mL}}\)

Step-by-step solution

To calculate the density of the \(_{12}^{24}Mg\)

We have simple formula from which we can calculate density:

\(\rho = \frac{m}{V}\)

The spherical shape has this formula for volume:

\(V = \frac{4}{3} \cdot \pi \cdot {r^3}\)

We have a diameter of\(1 \cdot {10^{ - 13}}\;{\rm{cm}}\), from which we can get\(r\):

\(\begin{aligned}{l}d &= 2r\\r &= \frac{1}{2} \cdot d\\r &= \frac{1}{2} \times 1 \times {10^{ - 13}}\;{\rm{cm}}\\r &= 5 \times {10^{ - 14}}\;{\rm{cm}}\end{aligned}\)

Now we need to get the mass of \(Na\)which is measured in amu (atomic mass unit):

\(1{\rm{amu}} = 1.67 \times {10^{ - 24}}\;{\rm{g}}\)

An atomic mass of Na is 24, which means

\(\begin{aligned}{c}24{\rm{amu}} &= 1.67 \times {10^{ - 24}}\;{\rm{g}} \times 24\\ &= 4.008 \times {10^{ - 23}}\;{\rm{g}}\end{aligned}\)

Now we can calculate the density:

\(\begin{aligned}{l}\rho &= \frac{m}{{\frac{4}{3}\pi {r^3}}}\\\rho &= \frac{{4.008 \times {{10}^{ - 23}}\;{\rm{g}}}}{{\frac{4}{3} \times \frac{{22}}{7} \times {{\left( {5 \times {{10}^{ - 14}}\;{\rm{cm}}} \right)}^3}}}\\\rho &= 7.66 \times {10^{16}}{\rm{g/c}}{{\rm{m}}^3}\end{aligned}\)

And since\({\rm{1c}}{{\rm{m}}^3} = 1{\rm{mL}}\),

\(\rho = 7.66 \times {10^{16}}{\rm{g/ml}}\)

Finally we get,

\(\rho = 7.66 \times {10^{16}}{\rm{g/mL}}\)