Modern Physics
Modern Physics
3rd Edition
ISBN: 9781111794378
Author: Raymond A. Serway, Clement J. Moses, Curt A. Moyer
Publisher: Cengage Learning
Question
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Chapter 12, Problem 11P

(a)

To determine

Show that the current density J=neμnE.

(a)

Expert Solution
Check Mark

Answer to Problem 11P

The current density J=neμnE is shown here.

Explanation of Solution

Write the expression for the current density,

    J=nevd        (I)

Here, J is the current density, n is the electron per unit volume, e is the charge of electron and vd is the drift velocity.

Write the expression for the drift velocity,

    vd=eτEme        (II)

Here, E is the electric field, τ is the average time and me is the mass of electron.

Write the expression for the drift velocity,

    vd=μnE        (III)

Here, μn is the mobility of the electrons.

Compare (II) and (III),

    μnE=eτEmeμn=eτme        (IV)

Conclusion:

Substitute (II) and (IV) in (I),

    J=ne2τEme=neμnE        (V)

Therefore, the current density J=neμnE is shown here.

(b)

To determine

Show that the both electrons and holes are present when σ=neμn+peμp .

(b)

Expert Solution
Check Mark

Answer to Problem 11P

The both electrons and holes are present when σ=neμn+peμp is shown here.

Explanation of Solution

Write the expression for the current density,

    J=σE        (VI)

Here, σ is the electrical conductivity.

Write the expression for the current density,

    J=Jelectrons+Jholes        (VII)

Here, Jelectrons is the current density of the electrons and Jholes is the current density of the holes.

Write the expression for the current density of the electrons,

    Jelectrons=neμnE        (VIII)

Here, n is the electron concentration.

Write the expression for the current density of the holes,

    Jelectrons=peμpE        (IX)

Here, p is the hole concentration.

Conclusion:

Substitute (VII), (VIII) and (IX) in (VI),

    σE=neμnE+peμpEσ=neμn+peμp

Therefore, the both electrons and holes are present when σ=neμn+peμp is shown here.

(c)

To determine

The drift speed of an electron .

(c)

Expert Solution
Check Mark

Answer to Problem 11P

The drift speed of an electron is 3.9×105cm/s_.

Explanation of Solution

From part (a), the equation (III)

Write the expression for the drift speed of an electron,

    vd=μnE

Here, μn is the mobility of the electrons.

Conclusion:

Substitute 3900cm2/Vs for μn and 100V/cm for E in (III),

    vd=(3900cm2/Vs)(100V/cm)=3.9×105cm/s

Therefore, the drift speed of an electron is 3.9×105cm/s_.

(d)

To determine

The resistivity of sample.

(d)

Expert Solution
Check Mark

Answer to Problem 11P

The resistivity of sample is 0.36Ωm_.

Explanation of Solution

From part (b), take the equation (X)

Write the expression for the electrical conductivity (n=p),

    σ=neμn+peμp=pe(μn+μp)        (X)

Write the expression for the resistivity of sample,

    ρ=1σ        (XI)

Here, ρ is the resistivity of sample.

Conclusion:

Substitute 3.0×1013cm-3 for p, 1.6×1019C for e and 5800cm2/Vs for μn+μp in (X),

    σ=(3.0×1013cm-3)(1.6×1019C)(5800cm2/Vs)=0.028A/Vcm=0.028(Ωcm)1=2.8(Ωm)1

Substitute 2.8(Ωm)1 for σ in (XI),

    ρ=12.8(Ωm)1=0.36Ωm

Therefore, the resistivity of sample is 0.36Ωm_.

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