Explanation:
(a). First of all, we calculate the number of atoms per cm^{3}, and by taking into account that each atom contributes with one free electron we have
[tex]n=\frac{\rho N_{A}}{M}=\frac{8.96\frac{g}{cm^{3}}*6.023*10^{23}\frac{atm}{mol}}{63.5\frac{g}{mol}}=8.49*10^{22}\frac{atm}{cm^{3}}[/tex][tex]\rho=n|e|=(6.02*10^{22}\frac{atm}{cm^{3}})(1.6*10^{-19}C)=96320\frac{C}{cm^{3}}[/tex]
(b)
[tex]v_{d}=\frac{J}{ne}=\frac{I}{Ane}=\frac{3.00A}{\pi (1.628*10^{-1}cm)^{2}(13600\frac{C}{cm^{3}})}=4.3*10^{-4}\frac{cm}{s}[/tex]
(c)
[tex]n=\frac{\rho N_{A}}{M}=\frac{2.70\frac{g}{cm^{3}}*6.023*10^{23}\frac{atm}{mol}}{26.98\frac{g}{mol}}=6.02*10^{22}\frac{atm}{cm^{3}}[/tex]
[tex]\rho=n|e|=(8.49*10^{22}\frac{atm}{cm^{3}})(1.6*10^{-19}C)=13600\frac{C}{cm^{3}}[/tex]
[tex]v_{d}=\frac{J}{ne}=\frac{I}{Ane}=\frac{3.00A}{\pi (1.628*10^{-1}cm)^{2}(96320\frac{C}{cm^{3}})}=6.08*10^{-5}\frac{cm}{s}[/tex]
I hope this is useful for you
P.D please change 1.628 (the diameter) by 0.814(the radius) in the calculation
:)
