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#conductivity

37 public questions tagged with this topic.

Semiconductors, Types and Energy Bands

The conductivity of an extrinsic semiconductor increases due to:

**Types of semiconductors** elemental Si, Ge group IV with 4 valence electrons, compound GaAs, InP etc. Intrinsic has n_e = n_h, extrinsic doped with pentavalent donors (P, As) gives n-type excess electrons, trivalent acceptors (B, Al) gives p-type excess holes, resistivity range semiconductors 10⁻⁵ to 10⁶ Ω·m vs insulators 10¹¹ Ω·m. Doping introduces impurities (pentavalent or trivalent) that provide additional charge carriers (electrons or holes), significantly enhancing conductivity compared to intrinsic semiconductors. Substituting values gives Addition of impurities, which matches expecte

Ref: NCERT > Physics Book > Electronic Devices > Semiconductors, Types and Energy Bands

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Semiconductors, Types and Energy Bands

The electrical conductivity of a semiconductor is:

**Types of semiconductors** elemental Si, Ge group IV with 4 valence electrons, compound GaAs, InP etc. Intrinsic has n_e = n_h, extrinsic doped with pentavalent donors (P, As) gives n-type excess electrons, trivalent acceptors (B, Al) gives p-type excess holes, resistivity range semiconductors 10⁻⁵ to 10⁶ Ω·m vs insulators 10¹¹ Ω·m. Semiconductors have intermediate conductivity ( 10⁵ to 10⁻⁶ S m⁻¹ ), between metals ( 10² to 10⁸ S m⁻¹ ) and insulators ( 10⁻¹¹ to 10⁻¹⁹ S m⁻¹ ). Substituting values gives Intermediate to metals and insulators, which matches expected behaviour for this semiconduct

Ref: NCERT > Physics Book > Electronic Devices > Semiconductors, Types and Energy Bands

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Electric Current, Drift Velocity and Mobility

In a conductor, if the number of free electrons per unit volume doubles while the electric field and relaxation time rem

**Current and drift relation** I = n e A v_d shows current proportional to drift velocity and area. For A=6×10⁻⁷ m², I=1.8 A, n=8.5×10²⁸ m⁻³, v_d =1.8/(8.5×10²⁸×1.6×10⁻¹⁹×6×10⁻⁷)=2.2×10⁻⁴ m/s, illustrating small drift speed even for ampere currents. Current density j = n e v_d , where v_d = e E tau / m . If n doubles and E , tau , and m remain constant, v_d is unchanged, so j' = 2n e v_d = 2j . Thus, current density doubles. Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0, R_eq series/parallel, V = ε - I

Ref: NCERT > Physics Book > Current Electricity > Electric Current, Drift Velocity and Mobility

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Electric Current, Drift Velocity and Mobility

What is the significance of the relaxation time in the context of electron drift in a conductor?

**Mobility** μ = v_d/E = e τ/m, τ relaxation time (s), measures ease of electron drift under field E (V/m). Conductivity σ = n e μ = 1/ρ, linking microscopic τ to macroscopic resistivity, explaining why metals conduct well due to large n and τ. Relaxation time ( tau ) is the average time between electron collisions with lattice ions. It determines drift velocity ( v_d = e E tau / m ), affecting how quickly electrons respond to the field and thus the conductor’s conductivity. Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0, R_eq series/parallel,

Ref: NCERT > Physics Book > Current Electricity > Electric Current, Drift Velocity and Mobility

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CHEMISTRY

What is the ll constant if the resistance of a conductivity ll with 0.01 M KCl solution is 200 Ω and its conductivity i

Given: What is the ll constant if the resistance of a conductivity ll with 0.01 M KCl solution is 200 Ω and its conductivity is 0.14 × 10⁻² S cm⁻¹? These values define the system as per NCERT data. Formula: Cell constant, G^* = kappa × R = 0.14 × 10⁻² × 200 = 0.28 cm^{-1 .. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Substituting values like 1.2 × 10⁻⁵, 236 J kg⁻¹ K⁻¹, CH₃CH₂NH₂ etc. into the formula and simplifying step by step. Result: The computed value matches the expected outcome and confirms the correc

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Solutions, Topic: Concentration terms and colligative properties.

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