Question

Three materials A, B, C have electrical conductivities σ, 2σ,and 2σ respectively their number densities of free electron are n, 2n and n respectively. for which material is the average collision time of free electrons maximum. ​

Answer 1

Answer:  

collision time for material B is maximum

Explanation:

Given: materials have electrical conductivities

A= σ , B=2σ  ,C=2σ

Densities of free electron : n , 2n , n

Respectively,

To find :  average collision time of free electrons maximum.

conductivity  σ=1ρ=ne2τm

Relaxation time, τ=mσne2. i.e., τ∝σn

∴τA:τB:τC=σ2n:2σn:2σ2n=σ2n:2σn:σn

Thus, τB>τC>τA. So average collision time for material B is maximum.

COLLISION TIME= At the individual particle level, the collision time is the mean time required for the direction of motion of an individual type particle to deviate through approximately as a consequence of collisions with particles of type.

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Answer 2

Answer:

Average collision time for material C is maximum

Explanation:

Conductivity is defined as the material which can show the ability to pass the current through it.

σ = 1/ρ

ρ=ne^2Tm

Relaxation time is defined as the time interval between two successive collisions of electrons in a conductor when the current flow.

T=mσne^2.

i.e., T∝σn

∴TA:TB:TC = σ/ n : 2σ/2n : 2σ/n

  TA:TB:TC =σ/n : σ/n : 2σ/n

Thus, TC>TA = TB. So average collision time for material C is maximum.

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