Question

For an iron nucleus with mass number \[A = 56\] and mass \[55.85\,\text{u}\], taking \[R_0 = 1.2 \times 10^{-15}\,\text{m}\] and \[R = R_0 A^{1/3}\], the calculated nuclear density using \[\rho = \frac{m}{V}\] is about:

Options

• \[1.2 \times 10^{15}\,\text{kg m}^{-3}\]

• \[5.585 \times 10^{17}\,\text{kg m}^{-3}\]

• \[2.29 \times 10^{7}\,\text{kg m}^{-3}\]

• \[2.29 \times 10^{17}\,\text{kg m}^{-3}\]

Answer 1

First find \[R \approx 4.6 \times 10^{-15}\,\text{m}\], then \[V = \frac{4}{3}\pi R^3 \approx 4.05 \times 10^{-43}\,\text{m}^3\]. Converting the mass, \[m = 55.85\,\text{u} \approx 9.27 \times 10^{-26}\,\text{kg}\]. Thus \[\rho = m/V \approx 2.29 \times 10^{17}\,\text{kg m}^{-3}\], consistent with the nearly constant nuclear density.