Chapter 12, Problem 12P

(a)

To determine

The drift speed of the electrons in GaAs.

Answer to Problem 12P

The drift speed of the electrons in GaAs is $85000\text{ cm/s}$ .

Explanation of Solution

Write the equation for the drift speed.

  $v_d=\mu E$        (I)

Here, $v_d$ is the drift speed, $\mu$ is the electron mobility and $E$ is the electric field.

Conclusion:

Substitute $8500{\text{ cm}}^2\text{/V}\cdot \text{s}$ for $\mu$ and $10\text{ V/cm}$ for $E$ in equation (I) to find $v_d$ .

  $\begin{array}{c}{v}_d=\left(8500{\text{ cm}}^2\text{/V}\cdot \text{s}\right)\left(10\text{ V/cm}\right)\\ =85000\text{ cm/s}\end{array}$

Therefore, the drift speed of the electrons in GaAs is $85000\text{ cm/s}$ .

(b)

To determine

The percent of the drift speed to the electron’s thermal speed at $300\text{ K}$ .

Answer to Problem 12P

The percent of the drift speed to the electron’s thermal speed at $300\text{ K}$ is $0.73\%$ .

Explanation of Solution

Write the equation connecting the kinetic energy and the thermal energy of the electron.

  $\frac{1}{2}m{v}_{\text{th}}^2=\frac{3k_{B}T}{2}$

Here, $m$ is the mass of the electron, $v_{\text{th}}$ is the thermal speed of the electron, $k_B$ is the Boltzmann constant and $T$ is the absolute temperature.

Rewrite the above equation for $v_{\text{th}}$ .

  $\begin{array}{c}m{v}_{\text{th}}^2=3k_{B}T\\ v_{\text{th}}^2=\frac{3k_{B}T}{m}\\ v_{\text{th}}=\sqrt{\frac{3k_{B}T}{m}}\end{array}$        (II)

Write the equation for the percentage of the drift speed to the electron’s thermal speed.

  $\text{Percentage}=\frac{v_d}{v_{\text{th}}}\times 100\%$        (III)

Conclusion:

The value of $k_B$ is $1.40\times {10}^{-23}\text{ J/K}$ and the mass of the electron is $9.11\times {10}^{-31}\text{ kg}$ .

Substitute $1.40\times {10}^{-23}\text{ J/K}$ for $k_B$ , $300\text{ K}$ for $T$ and $9.11\times {10}^{-31}\text{ kg}$ for $m$ in equation (II) to find $v_{\text{th}}$ .

  $\begin{array}{c}{v}_{\text{th}}=\sqrt{\frac{3\left(1.40\times {10}^{-23}\text{ J/K}\right)\left(300\text{ K}\right)}{9.11\times {10}^{-31}\text{ kg}}}\\ =117000\text{ m/s}\frac{100\text{ cm}}{1\text{ m}}\\ =1.17\times {10}^7\text{ cm/s}\end{array}$

Substitute $85000\text{ cm/s}$ for $v_d$ and $1.17\times {10}^7\text{ cm/s}$ for $v_{\text{th}}$ in equation (III) to find the percentage.

  $\begin{array}{c}\text{Percentage}=\frac{85000\text{ cm/s}}{1.17\times {10}^7\text{ cm/s}}\times 100\%\\ =0.73\%\end{array}$

Therefore, the percent of the drift speed to the electron’s thermal speed at $300\text{ K}$ is $0.73\%$ .

(c)

To determine

The average time between electron collisions.

Answer to Problem 12P

The average time between electron collisions is $4.8\times {10}^{-12}\text{ s}$ .

Explanation of Solution

Rewrite equation (I) for $\mu$ .

  $\mu =\frac{v_d}{E}$

Write the equation for the electron mobility.

  $\mu =\frac{e\tau }{m}$

Here, $e$ is the magnitude of the charge of the electron and $\tau$ is the average time between electron collisions.

Equate the above two equations and rewrite it for $\tau$ .

  $\begin{array}{c}\frac{e\tau }{m}=\frac{v_d}{E}\\ e\tau E=m{v}_d\\ \tau =\frac{m{v}_d}{eE}\end{array}$        (IV)

Conclusion:

The value of $e$ is $1.60\times {10}^{-19}\text{ C}$ .

Substitute $9.11\times {10}^{-31}\text{ kg}$ for $m$ , $85000\text{ cm/s}$ for $v_d$ , $1.60\times {10}^{-19}\text{ C}$ for $e$ and $10\text{ V/cm}$ for $E$ in equation (IV) to find $\tau$ .

  $\begin{array}{c}\tau =\frac{\left(9.11\times {10}^{-31}\text{ kg}\right)\left(85000\text{ cm/s}\right)}{\left(1.60\times {10}^{-19}\text{ C}\right)\left(10\text{ V/cm}\right)}\\ =4.8\times {10}^{-8}\text{ kg}\cdot {\text{cm}}^2\text{/J}\cdot \text{s}\frac{1{\text{ m}}^2}{{10}^4{\text{ cm}}^2}\\ =4.8\times {10}^{-12}\text{ s}\end{array}$

Therefore, the average time between electron collisions is $4.8\times {10}^{-12}\text{ s}$ .

(d)

To determine

The electronic mean path.

Answer to Problem 12P

The electronic mean path is $5600\stackrel{\circ }{\text{A}}$ .

Explanation of Solution

Write the equation for the electronic mean path.

  $L=v_{\text{th}}\tau$        (V)

Here, $L$ is the electronic mean path.

Conclusion:

Substitute $1.17\times {10}^7\text{ cm/s}$ for $v_{\text{th}}$ and $4.8\times {10}^{-12}\text{ s}$ for $\tau$ in equation (V) to find $L$ .

  $\begin{array}{c}L=\left(1.17\times {10}^7\text{ cm/s}\right)\left(4.8\times {10}^{-12}\text{ s}\right)\\ =5.6\times {10}^{-5}\text{ cm}\frac{1\stackrel{\circ }{\text{A}}}{{10}^{-8}\text{ cm}}\\ =5600\stackrel{\circ }{\text{A}}\end{array}$

Therefore, the electronic mean path is $5600\stackrel{\circ }{\text{A}}$ .