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When 5 V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons is 2.5×10-4ms-1 If the electron density in the wire is 8×1028m-3s, the resistivity of the material is close to
  • a)
     1.6×10-8Ωm
  • b)
     1.6×10​-7Ωm
  • c)
     1.6×10​-6Ωm
  • d)
     1.6×10​-5Ωm
Correct answer is option 'D'. Can you explain this answer?
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When 5 V potential difference is applied across a wire of length 0.1 m...
Given Data:
- Potential difference (V) = 5 V
- Length of wire (L) = 0.1 m
- Drift speed of electrons (v_d) = 2.5×10^-4 m/s
- Electron density (n) = 8×10^28 m^-3

Formula:
The resistivity (ρ) of a material can be calculated using the formula:
ρ = (m / n * e^2 * τ)
where,
m = electron mass = 9.1×10^-31 kg
e = charge of an electron = 1.6×10^-19 C
τ = relaxation time

Calculation:
Given drift speed is related to relaxation time by the formula:
v_d = e * E * τ / m
where,
E = electric field
Electric field (E) can be calculated using:
E = V / L
Substitute the values to find E:
E = 5V / 0.1m = 50 V/m
Now, substitute E and v_d in the drift speed formula to find τ:
2.5×10^-4 = 1.6×10^-19 * 50 * τ / 9.1×10^-31
τ = 9.1×10^-31 * 2.5×10^-4 / (1.6×10^-19 * 50)
τ ≈ 2.273×10^-14 s
Finally, substitute all the values in the resistivity formula:
ρ = (9.1×10^-31 / (8×10^28 * (1.6×10^-19)^2 * 2.273×10^-14)
ρ ≈ 1.6×10^-5 Ωm
Therefore, the resistivity of the material is close to 1.6×10^-5 Ωm, which corresponds to option 'D'.
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When 5 V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons is2.5×10-4ms-1If the electron density in the wire is8×1028m-3s, the resistivity of the material is close toa)1.6×10-8Ωmb)1.6×10-7Ωmc)1.6×10-6Ωmd)1.6×10-5ΩmCorrect answer is option 'D'. Can you explain this answer?
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