Question 42.7: What is the nuclear mass density ρm of (a) the fairly low-ma......
What is the nuclear mass density ρ_m of (a) the fairly low-mass nuclide ^{55}Mn and (b) the fairly high-mass nuclide ^{209}Bi? (c) Compare the two answers, with an explanation. What is the nuclear charge density ρ_q of (d) ^{55}Mn and (e) ^{209}Bi? (f) Compare the two answers, with an explanation.
For ^{55}Mn the mass density is
\rho_m=\frac{M}{V}=\frac{0.055 \,kg / mol }{(4 \pi / 3)\left[\left(1.2 \times 10^{-15}\, m \right)(55)^{1 / 3}\right]^3\left(6.02 \times 10^{23} / mol \right)}=2.3 \times 10^{17} \,kg / m ^3 .
(b) For ^{209}Bi,
\rho_m=\frac{M}{V}=\frac{0.209 \mathrm{~kg} / \mathrm{mol}}{(4 \pi / 3)\left[\left(1.2 \times 10^{-15} \mathrm{~m}\right)(209)^{1 / 3}\right]^3\left(6.02 \times 10^{23} / \mathrm{mol}\right)}=2.3 \times 10^{17} \mathrm{~kg} / \mathrm{m}^3 .
(c) Since V \propto r^3=\left(r_0 A^{1 / 3}\right)^3 \propto A, we expect \rho_m \propto A / V \propto A / A \approx \text { const } . for all nuclides.
(d) For ^{55}Mn, the charge density is
\rho_q=\frac{Z e}{V}=\frac{(25)\left(1.6 \times 10^{-19} \mathrm{C}\right)}{(4 \pi / 3)\left[\left(1.2 \times 10^{-15} \mathrm{~m}\right)(55)^{1 / 3}\right]^3}=1.0 \times 10^{25} \mathrm{C} / \mathrm{m}^3
(e) For ^{209}Bi, the charge density is
\rho_q=\frac{Z e}{V}=\frac{(83)\left(1.6 \times 10^{-19} \mathrm{C}\right)}{(4 \pi / 3)\left[\left(1.2 \times 10^{-15} \mathrm{~m}\right)(209)^{1 / 3}\right]^3}=8.8 \times 10^{24} \mathrm{C} / \mathrm{m}^3.
Note that \rho_q \propto Z / V \propto Z / A A should gradually decrease since A > 2Z for large nuclides.