Q.How does the mobility of electrons in a conductor change, if the potential difference applied across the conductor is doubled, keeping the length and temperature of the conductor constant?

Showing the 12 most recent of 13 on this concept.

1. CBSE 2026Set ANNUAL1 mark
   Q.What is the S.I. unit of mobility (mu)?

   ›Reveal solutionSolution

   Mobility is drift velocity per unit electric field, so its unit works out to m^2/(V.s).

   Mobility mu of a charge carrier is defined as the drift velocity acquired per unit applied electric field:

   mu = vd / E

   vd (drift velocity) has SI unit m/s, and E (electric field) has SI unit V/m. So

   unit of mu = (m/s) / (V/m) = m^2 / (V.s) = m^2 V^-1 s^-1

   ✓Final answer

   m^2 V^-1 s^-1 (square metre per volt-second).

2. CBSE 2025Set ANNUAL1 mark
   Q.Magnitude of the drift velocity per unit electric field is equal to ____________.

   ›Reveal solutionSolution

   Drift velocity per unit applied field is, by definition, the mobility of the charge carriers.

   When an electric field $E$ is applied across a conductor, free charge carriers acquire an average drift velocity $v_d$ proportional to $E$ (for ohmic conductors):

   $v_d = \mu E \quad \Rightarrow \quad \mu = \dfrac{v_d}{E}$

   Here $\mu$ is called the mobility of the charge carrier, with SI unit m$^2$V$^{-1}$s$^{-1}$.

   ✓Final answer

   Magnitude of the drift velocity per unit electric field is equal to the mobility ($\mu$) of the charge carriers.

3. CBSE 2025Set ANNUAL1 mark
   Q.How does the mobility of electron in a conductor change, if the potential difference applied across the conductor is double keeping the length and temperature of the conductor constant?

   ›Reveal solutionSolution

   Mobility $\mu = e\tau/m$ depends only on the material, not on the applied field; $v_d=\mu E$ changes, $\mu$ does not.

   Mobility is defined as drift velocity per unit electric field, $\mu = v_d/E = e\tau/m$, which depends only on the electron charge, mass, and the relaxation time $\tau$ (a property of the material at a given temperature) — it does not depend on the strength of the applied field. If the potential difference (and hence the field $E$, with length fixed) is doubled, the drift velocity $v_d = \mu E$ doubles proportionally, but the mobility $\mu$ itself remains exactly the same, since temperature (and hence $\tau$) is unchanged.

   ✓Final answer

   Mobility is unchanged (it depends only on $\tau$, $e$, $m$ — a material property at fixed temperature); it is the drift velocity that doubles.

4. CBSE 2024Set ANNUAL1 markMCQ
   Q.The SI unit of mobility ($\mu$) of charge carriers is -

   (a) $m^2V^{-1}$

   (b) $m^2V^{-1}s^{-1}$

   (c) $m^2Vs$

   (d) $m^2V^2s$

   ›Reveal solutionSolution

   Mobility is drift velocity per unit electric field, so its unit works out to $m^2V^{-1}s^{-1}$.

   Mobility $\mu = v_d/E$, where $v_d$ is drift velocity (unit $ms^{-1}$) and $E$ is electric field (unit $Vm^{-1}$). So the SI unit of $\mu$ is $\dfrac{ms^{-1}}{Vm^{-1}} = m^2V^{-1}s^{-1}$.

   ✓Final answer

   (b) $m^2V^{-1}s^{-1}$

5. CBSE 2024Set ANNUAL1 mark
   Q.The magnitude of drift velocity of electron per unit electric field is called __________.

   ›Reveal solutionSolution

   Mobility is defined precisely as drift speed per unit electric field, and characterises how easily charge carriers drift under a field.

   When an electric field E is applied across a conductor, free electrons acquire an average drift velocity $v_d$ superimposed on their random thermal motion. The quantity:

   $\mu = \dfrac{v_d}{E}$

   is called the mobility of the electron. It measures how effectively the field can drift the charge carriers, and has SI unit m²/(V·s).

   ✓Final answer

   Mobility.

6. CBSE 2023Set ANNUAL1 markMCQ
   Q.A charged particle has drift velocity 4.2 x 10^-4 m s^-1 in electric field of 2.1 x 10^-10 V m^-1. Its mobility is

   (1) 0.5 x 10^6 m^2 V^-1 s^-1

   (2) 2 x 10^6 m^2 V^-1 s^-1

   (3) 0.5 x 10^-6 m^2 V^-1 s^-1

   (4) 2 x 10^-6 m^2 V^-1 s^-1

   ›Reveal solutionSolution

   Mobility is the drift velocity acquired per unit electric field.

   $\mu=\dfrac{v_d}{E}=\dfrac{4.2\times10^{-4}}{2.1\times10^{-10}}=2\times10^{6}\ \text{m}^2\text{V}^{-1}\text{s}^{-1}$

   ✓Final answer

   (2) $2\times10^{6}\ \text{m}^2\text{V}^{-1}\text{s}^{-1}$.

7. CBSE 2022Set GC1 mark
   Q.Define mobility.

   ›Reveal solutionSolution

   Mobility $\mu=v_d/E$ — drift velocity per unit applied electric field.

   Definition. The mobility of a charge carrier is the magnitude of the drift velocity it acquires per unit electric field applied across the conductor:

   $\mu=\frac{v_d}{E}.$

   Since the drift velocity is $v_d=\dfrac{eE\tau}{m}$, we also have $\mu=\dfrac{e\tau}{m}$. Its SI unit is $\text{m}^2\,\text{V}^{-1}\text{s}^{-1}$. A larger mobility means the carriers respond more readily to a field, giving higher conductivity.

   ✓Final answer

   $\mu=\dfrac{v_d}{E}=\dfrac{e\tau}{m}$, units $\text{m}^2\,\text{V}^{-1}\text{s}^{-1}$.

8. CBSE 2022Set I1 markMCQ
   Q.Which of the following is correct for mobility? (A) μ = V_d/E (B) μ = E/V_d (C) μ = V_d·E (D) μ = E^2·V_d

   ›Reveal solutionSolution

   Mobility μ = drift velocity per unit field = V_d/E.

   Mobility measures how readily charge carriers drift in an applied electric field. It is defined as the magnitude of drift velocity per unit electric field:

   $\mu = \dfrac{v_d}{E}$

   Its SI unit is m²V⁻¹s⁻¹. A larger mobility means the carriers move faster for the same field.

   ✓Final answer

   (A) μ = V_d/E.

9. CBSE 2022Set ANNUAL1 mark
   Q.Magnitude of the drift velocity per unit electric field is ____.

   ›Reveal solutionSolution

   The drift velocity acquired per unit applied electric field is the mobility of the charge carrier.

   When an electric field E is applied across a conductor, free electrons acquire a small average velocity along the field called the drift velocity $v_d$. The magnitude of drift velocity per unit electric field, $\mu = v_d/E$, is defined as the mobility of the charge carrier. Its SI unit is $m^2 V^{-1}s^{-1}$. Mobility is related to conductivity by $\sigma = ne\mu$, where n is the number density of charge carriers and e is the electronic charge.

   ✓Final answer

   This quantity is called mobility, $\mu = v_d/E$.

10. CBSE 2020Set 55/3/11 markMCQ
    Q.$\mathrm{m^2\,V^{-1}\,s^{-1}}$ is the SI unit of which of the following ? (A) Drift velocity (B) Mobility (C) Resistivity (D) Potential gradient

    ›Reveal solutionSolution

    The unit $\mathrm{m^2\,V^{-1}\,s^{-1}}$ describes how much velocity a charge carrier gains per unit electric field — that's the definition of mobility. The answer is (B).

    When a charge carrier moves through a conductor under an electric field, two quantities matter: how fast it drifts (drift velocity) and how responsive it is to the field (mobility). The unit $\mathrm{m^2\,V^{-1}\,s^{-1}}$ tells us we're measuring a velocity-per-field ratio, which is precisely what mobility captures.

    Let's check each option systematically by dimensional analysis.

    1. Drift velocity is simply a speed — the average velocity of charge carriers in a conductor when an electric field is applied. Its SI unit is $\mathrm{m\,s^{-1}}$, not $\mathrm{m^2\,V^{-1}\,s^{-1}}$. So (A) is out.

    2. Mobility $\mu$ is defined as the drift velocity $v_d$ acquired per unit electric field $E$:

    $\mu = \frac{v_d}{E}$

    The electric field $E$ has units $\mathrm{V\,m^{-1}}$ (voltage per distance). Therefore:

    $[\mu] = \frac{\mathrm{m\,s^{-1}}}{\mathrm{V\,m^{-1}}} = \frac{\mathrm{m\,s^{-1}} \cdot \mathrm{m}}{\mathrm{V}} = \mathrm{m^2\,V^{-1}\,s^{-1}}$

    This matches exactly.

    3. Resistivity $\rho$ relates the electric field to current density via $E = \rho J$. Since current density $J$ has units $\mathrm{A\,m^{-2}}$ and $E$ has units $\mathrm{V\,m^{-1}}$:

    $[\rho] = \frac{\mathrm{V\,m^{-1}}}{\mathrm{A\,m^{-2}}} = \mathrm{V\,m\,A^{-1}} = \Omega\,\mathrm{m}$

    Not a match.

    4. Potential gradient is just another name for electric field strength, so its unit is $\mathrm{V\,m^{-1}}$. Also not a match.

    Tip

    Mobility is the "ease of motion" of a charge carrier in a material. High mobility means carriers respond strongly to even weak fields — crucial in semiconductor physics and device design.

    ✓Final answer

    The correct option is (B) Mobility.

11. CBSE 2019Set 55/1/11 mark
    Q.How does the mobility of electrons in a conductor change, if the potential difference applied across the conductor is doubled, keeping the length and temperature of the conductor constant?

    ›Reveal solutionSolution

    Mobility of electrons in a conductor is independent of the applied potential difference — it depends only on the material and temperature. Doubling the voltage does not change mobility.

    The Concept: What Mobility Really Means

    Mobility ($\mu$) is a material property that tells us how fast an electron can drift through a conductor when an electric field is applied. It is defined as:

    $\mu = \frac{v_d}{E}$

    where $v_d$ is the drift velocity of electrons and $E$ is the electric field inside the conductor.

    The key insight: mobility is not a measure of how many electrons move, but how easily they move through the lattice. It depends on:

    • The nature of the material (atomic structure, number of free electrons)
    • Temperature (which affects lattice vibrations and scattering)

    It does not depend on the applied voltage or current — those are external conditions.

    Step-by-Step Reasoning

    1. What happens when voltage is doubled?

       For a conductor of fixed length $L$, the electric field is $E = V/L$. Doubling $V$ doubles $E$.

    2. How does drift velocity respond?

       Drift velocity is proportional to the electric field: $v_d \propto E$. So when $E$ doubles, $v_d$ also doubles. This is because a stronger field exerts a larger force on each electron ($F = eE$), accelerating it more between collisions.

    3. Now look at the mobility formula:

       $\mu = v_d / E$. If both $v_d$ and $E$ double, their ratio stays exactly the same.

       Mathematically:

    $\mu_{\text{new}} = \frac{2v_d}{2E} = \frac{v_d}{E} = \mu_{\text{old}}$

    4. The deeper reason — why mobility is constant here: Mobility is determined by the average time between collisions ($\tau$) of electrons with lattice ions:

    $\mu = \frac{e\tau}{m}$

    where $e$ is electron charge and $m$ is electron mass.

    The relaxation time $\tau$ depends on temperature (which is fixed here) and the lattice structure — not on the applied field. So $\mu$ cannot change.

    Watch out

    A common mistake is to think that since current increases with voltage, mobility must also increase. But current depends on both drift velocity and charge carrier density ($I = n e A v_d$). The increase in current comes entirely from the increase in $v_d$, not from a change in $\mu$.

    Tip

    Think of mobility like the "ease of sliding" for electrons through the material. Doubling the voltage is like pushing harder — they slide faster, but the slipperiness of the surface (mobility) hasn't changed.

    ✓Final answer

    The mobility of electrons remains unchanged when the potential difference is doubled, because mobility depends only on the material and temperature, not on the applied voltage.

12. CBSE 2019Set ANNUAL1 mark
    Q.Write the unit of mobility.

    ›Reveal solutionSolution

    Mobility is drift velocity per unit electric field, so its unit is (m/s) divided by (V/m), giving m²V⁻¹s⁻¹.

    Mobility $\mu$ of a charge carrier is defined as the magnitude of drift velocity acquired per unit applied electric field:

    $\mu = \dfrac{v_d}{E}$

    Units: $v_d$ is in $\text{m s}^{-1}$, and $E$ (electric field) is in $\text{V m}^{-1}$. So:

    $[\mu] = \dfrac{\text{m s}^{-1}}{\text{V m}^{-1}} = \text{m}^2 \text{V}^{-1}\text{s}^{-1}$

    ✓Final answer

    Unit of mobility $= \text{m}^2\text{V}^{-1}\text{s}^{-1}$.