Physics · Current Electricity · NEET
No. Drift velocity v_d is the actual slow speed (in m/s) of a carrier in a field. Mobility μ is v_d divided by the field E, so μ = v_d/E. Mobility tells you how much drift you get per unit field, so two carriers can have the same drift velocity but different mobility if the fields are different.
The SI unit is m²/Vs (metre squared per volt-second). It comes from dividing drift velocity (m/s) by field (V/m): (m/s)/(V/m) = m²/(V·s). In practical work you may see cm²/Vs, which is 10^4 times smaller than the SI unit.
Mobility is defined as the magnitude of drift velocity per unit field. For a positive carrier the drift is along E; for an electron it is opposite to E. But because we take the magnitude, μ is positive for every carrier. So you never write a negative mobility in NEET answers.
For a normal metal at a fixed temperature, mobility is roughly a constant of the material because μ = eτ/m, and the relaxation time τ does not depend much on E. Increasing E increases the drift velocity, but v_d and E rise together so their ratio μ stays about the same. Mobility mainly changes with temperature, not with the applied field.
Since drift velocity v_d = eEτ/m, dividing by E gives μ = v_d/E = eτ/m. Here e is the carrier charge, m its mass, and τ the relaxation time (average time between collisions). So a longer relaxation time or a smaller mass means higher mobility.
In a semiconductor an electron moves freely, while a hole moves by electrons hopping into it, which is a slower, more hindered process. So μ_e > μ_h. For equal carrier concentration and equal field, the material with higher mobility (n-type) carries the larger current.
A charged particle having drift velocity 7.5×10^-4 m/s in an electric field of 3×10^-10 V/m has a mobility (in m²/Vs) of:
The electron concentration in an n-type semiconductor equals the hole concentration in a p-type. An electric field is applied across each. Compare the currents:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Mobility μ is the drift velocity of a charge carrier per unit applied electric field, μ = v_d/E, measured in m²/Vs.
μ = v_d/E, and using v_d = eEτ/m it also equals μ = eτ/m, where τ is the relaxation time, e the charge and m the mass of the carrier.
m²/Vs (metre squared per volt-second). It is 10^4 times the practical unit cm²/Vs.
No. Mobility is defined using the magnitude of drift velocity, so it is always positive for any carrier, including electrons.
Current I = n e A μ E, so for the same carrier density and field, higher mobility means larger current. This is why n-type semiconductors conduct more than p-type at equal conditions.
Yes. In metals, rising temperature shortens the relaxation time τ, so mobility falls. Since μ = eτ/m, a smaller τ gives smaller μ.
