Correct Answer - Option 3 : 1.29 × 10
-4 m/s
The correct answer is option 3) i.e. 1.29 × 10-4 m/s
CONCEPT:
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Drift velocity: The average velocity attained by the electrons under the influence of an electric field is termed the drift velocity.
- The free electrons are always in a constant random motion in the conductor.
- When the conductor is connected to a battery, an electric field is generated.
- When these free electrons are subjected to an electric field, they slowly drift in the direction of the electric field applied, while maintaining the random motion.
- At this point, the net velocity of the electrons is called its drift velocity.
The drift velocity (vd) of the electron in a conductor is given by:
\(v_d =\frac{I}{nAq}\)
Where I is the current flowing through the conductor, A is the area of cross-section of the conductor, q is the charge on the electron and n is the number of electrons per unit volume.
CALCULATION:
Given that:
Cross-sectional area, A = 4 × 10-6 m2
Current = 5 A
Density of aluminium = 2.7 × 103 kg/m3
Atomic weight of aluminium = 27 u = 27 g = 27 × 10-3 kg
Number of electrons per unit volume, \(n =\frac{No. \:of \:electrons}{Volume}\)
Here, as each atom provides one electron, the number of electrons is equal to the number of atoms.
Number of atoms in 27 u of aluminium = 6.02 × 1023 atoms
\(\Rightarrow n =\frac{No. \:of \:atoms}{Volume} = \frac{6.02 \;× \;10 ^{23}}{\frac{mass}{\rho}}\)
\(\Rightarrow n = \frac{6.02\; × \;10 ^{23}}{\frac{27\;× \;10^{-3} }{2.7 \;× \;10^{3}}} = 6.02 × 10^{28} \:electrons\: per \: m^3\)
\(\Rightarrow v_d =\frac{I}{nAq}\)
\(\Rightarrow v_d =\frac{5}{(6.02\; \times \;10^{28})\;\times\; (4\; \times\; 10^{-6}) \;\times \;(1.6 \;\times \;10^{-19})}\)
\(\Rightarrow v_d = 1.29 \times 10^{-4} \: m/s\)
