1-7 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} 1\ut{mol}&=6.02\times10^{23}\ut{EA}\\ \end{cases} $$ $$ \begin{aligned} N^3 &= 0.2\ut{mol}\\ N &= \sqrt[3]{0.2\times6.02\times10^{23}}\\ &\approx 4.93789854178\times10^7\\ &\approx 4.9\times10^7\ut{EA}\\ \end{aligned} $$ $$ \begin{aligned} \Ans&=4.9\times10^7\ut{cm}\\ &=4.9\times10^7\ut{cm}\times \cfrac{1\ut{m}}{100\ut{cm}}\times\cfrac{1\ut{km}}{1000\ut{m}}\\ &=4.9\times10^2\ut{km}\\ \end{aligned} $$