Chapter 12, Problem 11P

(a)

To determine

Show that the current density $J=ne{\mu }_{n}E$.

Answer to Problem 11P

The current density $J=ne{\mu }_{n}E$ is shown here.

Explanation of Solution

Write the expression for the current density,

    $J=ne{v}_d$        (I)

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

Write the expression for the drift velocity,

    $v_d=\frac{e\tau E}{m_e}$        (II)

Here, $E$ is the electric field, $\tau$ is the average time and $m_e$ is the mass of electron.

Write the expression for the drift velocity,

    $v_d={\mu }_{n}E$        (III)

Here, ${\mu }_n$ is the mobility of the electrons.

Compare (II) and (III),

    $\begin{array}{c}{\mu }_{n}E=\frac{e\tau E}{m_e}\\ {\mu }_n=\frac{e\tau }{m_e}\end{array}$        (IV)

Conclusion:

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

    $\begin{array}{c}J=\frac{n{e}^2\tau E}{m_e}\\ =ne{\mu }_{n}E\end{array}$        (V)

Therefore, the current density $J=ne{\mu }_{n}E$ is shown here.

(b)

To determine

Show that the both electrons and holes are present when $\sigma =ne{\mu }_n+pe{\mu }_p$ .

Answer to Problem 11P

The both electrons and holes are present when $\sigma =ne{\mu }_n+pe{\mu }_p$ is shown here.

Explanation of Solution

Write the expression for the current density,

    $J=\sigma E$        (VI)

Here, $\sigma$ is the electrical conductivity.

Write the expression for the current density,

    $J=J_{\text{electrons}}+J_{\text{holes}}$        (VII)

Here, $J_{\text{electrons}}$ is the current density of the electrons and $J_{\text{holes}}$ is the current density of the holes.

Write the expression for the current density of the electrons,

    $J_{\text{electrons}}=ne{\mu }_{n}E$        (VIII)

Here, $n$ is the electron concentration.

Write the expression for the current density of the holes,

    $J_{\text{electrons}}=pe{\mu }_{p}E$        (IX)

Here, $p$ is the hole concentration.

Conclusion:

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

    $\begin{array}{c}\sigma E=ne{\mu }_{n}E+pe{\mu }_{p}E\\ \sigma =ne{\mu }_n+pe{\mu }_p\end{array}$

Therefore, the both electrons and holes are present when $\sigma =ne{\mu }_n+pe{\mu }_p$ is shown here.

(c)

To determine

The drift speed of an electron .

Answer to Problem 11P

The drift speed of an electron is $\underset{\_}{3.9\times {10}^5\text{cm/s}}$.

Explanation of Solution

From part (a), the equation (III)

Write the expression for the drift speed of an electron,

    $v_d={\mu }_{n}E$

Here, ${\mu }_n$ is the mobility of the electrons.

Conclusion:

Substitute $3900{\text{cm}}^{\text{2}}\text{/Vs}$ for ${\mu }_n$ and $100\text{V/cm}$ for $E$ in (III),

    $\begin{array}{c}{v}_d=\left(3900{\text{cm}}^{\text{2}}\text{/Vs}\right)\left(100\text{V/cm}\right)\\ =3.9\times {10}^5\text{cm/s}\end{array}$

Therefore, the drift speed of an electron is $\underset{\_}{3.9\times {10}^5\text{cm/s}}$.

(d)

To determine

The resistivity of sample.

Answer to Problem 11P

The resistivity of sample is $\underset{\_}{0.36\Omega \text{m}}$.

Explanation of Solution

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

Write the expression for the electrical conductivity $\left(n=p\right)$,

    $\begin{array}{c}\sigma =ne{\mu }_n+pe{\mu }_p\\ =pe\left({\mu }_n+{\mu }_p\right)\end{array}$        (X)

Write the expression for the resistivity of sample,

    $\rho =\frac{1}{\sigma }$        (XI)

Here, $\rho$ is the resistivity of sample.

Conclusion:

Substitute $3.0\times {10}^{13}{\text{cm}}^{\text{-3}}$ for $p$, $1.6\times {10}^{-19}\text{C}$ for $e$ and $5800{\text{cm}}^2\text{/Vs}$ for ${\mu }_n+{\mu }_p$ in (X),

    $\begin{array}{c}\sigma =\left(3.0\times {10}^{13}{\text{cm}}^{\text{-3}}\right)\left(1.6\times {10}^{-19}\text{C}\right)\left(5800{\text{cm}}^2\text{/Vs}\right)\\ =0.028\text{A/V}\text{cm}\\ =0.028{\left(\Omega \text{cm}\right)}^{-1}\\ =2.8{\left(\Omega \text{m}\right)}^{-1}\end{array}$

Substitute $2.8{\left(\Omega \text{m}\right)}^{-1}$ for $\sigma$ in (XI),

    $\begin{array}{c}\rho =\frac{1}{2.8{\left(\Omega \text{m}\right)}^{-1}}\\ =0.36\Omega \text{m}\end{array}$

Therefore, the resistivity of sample is $\underset{\_}{0.36\Omega \text{m}}$.