Question

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

• A. $ 8.6 × 10^{−4}\, ms^{−1} $
• B. $ 1.3 × 10^{−4}\, ms^{−1} $
• C. $ 2.8 × 10^{−2}\, ms^{−1} $
• D. $ 3.8 × 10^{−3}\, ms^{−1} $

Answer 1

Answer: (B)

Mass of $1 \,m ^{3}$ 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 $27\, u$ , therefore, $27\, g$ 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=5 A , A=4 \times 10^{-6} m ^{2}, e=1.6 \times 10^{-19} C $
$\therefore v_{d}=\frac{I}{e n A} $
$= \frac{5}{1.6 \times 10^{-19} \times 6 \times 10^{28} \times 4 \times 10^{-6}}$
$= 1.3 \times 10^{-4} ms ^{-1} $