1-25 할리데이 10판 솔루션 일반물리학

(a) $$ \begin{cases} R &= 1.90\ut{\AAA} \\ N_0 &= 6.023 \times 10^{23}\ut{EA/mol} \\ m_{\mathrm{Na}} &= 22.9898\ut{g/mol} \end{cases} $$ $$ \begin{aligned} \Ans &= \rho_{\mathrm{Na}} = ? \\ &= \frac{\mathrm{Mass}}{\mathrm{Volume}} \\ &= \frac{m_{\mathrm{Na}}}{N_0} \cdot \frac{3}{4\pi R^3} \\ &= \frac{22.9898\ut{g/mol}}{6.023 \times 10^{23}\ut{EA/mol}} \cdot \frac{3}{4\pi (1.90\ut{\AAA})^3} \\ &= \frac{22.9898\ut{g/mol}}{6.023 \times 10^{23}\ut{EA/mol}} \cdot \frac{3}{4\pi (1.90\ut{\AAA})^3}\cdot \frac{1\ut{kg}}{1000\ut{g}}\cdot \left(\frac{10^{10}\ut{\AAA}}{1\ut{m}}\right)^3 \\ &= \frac{172423500000}{41311757 \pi}\ut{kg/m^3} \\ &\approx 1328.5347476363890999\ut{kg/m^3} \\ &\approx 1.33 \times 10^3 \ut{kg/m^3} \end{aligned} $$
(b) 결정상태에선 빈공간이 생겨서.