User:Jflm

${\frac {a}{b}}$

$\displaystyle \rho _{Cu}={\frac {4{\frac {m_{Cu}}{N_{A}}}}{a^{3}}}$

$\displaystyle \rho _{NaCl}={\frac {\frac {m_{Na}+m_{Cl}}{N_{A}}}{2d^{3}}}$

$\displaystyle \rho _{Cu}={\frac {m_{Cu}}{V_{Cu}}}={\frac {m_{Cu}}{\frac {m_{e/Cu}}{\rho _{e}}}}={\frac {m_{Cu}\rho _{e}}{m_{p+e}+m_{Cu}-m_{p+e+Cu}}}$