Question

An aluminium $(Al)$ rod with area of cross-section $4\times 10^{-6}{m}^2$ has a current of $5A$ flowing through it. Find the drift velocity of electron in the rod. Density of $Al=2.7\times 10^{3}kg{m}^{-3}$ and atomic $wt.=27u$ . Assume that each $Al$ atom provides one electron.

• A. $8.6\times 10^{-4}m{s}^{-1}$
• B. $1.3\times 10^{-4}m{s}^{-1}$
• C. $2.8\times 10^{-2}m{s}^{-1}$
• D. $3.8\times 10^{-3}m{s}^{-1}$

Answer 1

Answer: (B)

Mass of $1m^3$ of $Al$
$=2.7\times 10^{3}kg=2.7\times 10^{6}g$
Since Avogadro's number is $6.0\times 10^{23}$ and atomic mass of $Cl$ is $27u$ , therefore, $27g$ of $Al$ contains $6.0\times 10^{23}$ atoms.
So, $2.7\times 10^{6}g$ of $Al$ contains
$\frac{6.0\times 10^{23}}{27}\times 2.7\times 10^6$ atoms $=6\times 10^{28}$ atoms
Number of conduction electrons.
$n=$ number of Al atoms $=6\times 10^{28}$
Now, $I=5A,A=4\times 10^{-6}{m}^2,e=1.6\times 10^{-19}C$
$\therefore v_d=\frac{I}{enA}$
$=\frac{5}{1.6\times 10^{-19}\times 6\times 10^{28}\times 4\times 10^{-6}}$
$=1.3\times 10^{-4}m{s}^{-1}$