Chapter 14.12, Problem 75AAP

(a)

To determine

The concentration of majority carriers.

Answer to Problem 75AAP

The concentration of majority carriers is $6.17\times {10}^{18}\text{electrons}/{\text{cm}}^{\text{3}}$.

Explanation of Solution

Write the expression to calculate the concentration of majority charge carriers in a n-type semiconductor.

    $n_n=\frac{1}{\rho q{\mu }_n}$                                                                                       ...... (I)

Here, the concentration of majority carriers is $n_n$, charge  of electron is $q$ and the electron mobility is ${\mu }_n$.

Conclusion:

Refer to the Figure-14.26 (“The effect of total ionized impurity concentration on the mobility of charge carriers in silicon at room temperature.”) to obtain the value of electron mobility as $1350{\text{cm}}^2/\text{V}\cdot \text{s}$ at $27°\text{C}$.

Substitute $7.5\times {10}^{-4}\Omega \cdot \text{cm}$ for $\rho$, $1.60\times {10}^{-19}\text{C}$ for $q$ and $1350{\text{cm}}^2/\text{V}\cdot \text{s}$ in equation (I).

    $\begin{array}{c}{n}_n=\frac{1}{\left(7.5\times {10}^{-4}\Omega \cdot \text{cm}\right)\times \left(1.60\times {10}^{-19}\text{C}\right)\times \left(1350{\text{cm}}^2/\text{V}\cdot \text{s}\right)}\\ =\frac{1}{\left(7.5\times {10}^{-4}\Omega \cdot \text{cm}\right)\left(2160\times {10}^{-19}\text{C}\cdot {\text{cm}}^{\text{2}}/\text{V}\cdot \text{s}\right)}\\ =\frac{1}{16200\times {10}^{-23}{\text{cm}}^3}\\ =6.17\times {10}^{18}{\text{cm}}^{\text{-3}}\end{array}$

Thus, the concentration of majority carriers is $6.17\times {10}^{18}\text{electrons}/{\text{cm}}^{\text{3}}$.

(b)

To determine

The ratio of arsenic to silicon atoms.

Answer to Problem 75AAP

Thus the ratio of arsenic to silicon atoms is $1.24\times {10}^{-4}$.

Explanation of Solution

Write the expression to calculate silicon atoms.

    $N_{\text{Si}}=\frac{{\rho }_{\text{Si}}{N}_o}{\text{atomic}\text{weight}\text{of}\text{silicon}}$                                                 ...... (I)

Here, the number of silicon atoms is $N_{\text{Si}}$, the density of silicon is ${\rho }_{\text{Si}}$ and the Avogadro number is $N_o$.

Write the expression for ratio arsenic to the silicon atoms.

    $R=\frac{N_{\text{As}}}{N_{\text{Si}}}$                                                                                  ...... (II)

Here, the number of arsenic atoms is $N_{\text{As}}$.

Conclusion:

Substitute $2.33\text{g}/{\text{cm}}^{\text{3}}$ for ${\rho }_{\text{Si}}$, $6.023\times {10}^{23}\text{atoms}/\text{mol}$ for $N_o$ and $28.09\text{g}/\text{mol}$ for atomic weight of silicon in equation (I).

    $\begin{array}{c}{N}_{\text{Si}}=\frac{\left(2.33\text{g}/{\text{cm}}^{\text{3}}\right)\times \left(6.023\times {10}^{23}\text{atoms}/\text{mol}\right)}{28.09\text{g}/\text{mol}}\\ =\frac{14.03359\times {10}^{23}\text{g}\cdot \text{atoms}/{\text{cm}}^{\text{3}}\cdot \text{mol}}{28.09\text{g}/\text{mol}}\\ =4.99\times {10}^{22}\text{atoms}/{\text{cm}}^{\text{3}}\end{array}$

Substitute $4.99\times {10}^{22}\text{atoms}/{\text{cm}}^{\text{3}}$ for $N_{\text{Si}}$ and $6.17\times {10}^{18}\text{atoms}/{\text{cm}}^{\text{3}}$ for $N_{\text{As}}$ in equation (II).

    $\begin{array}{c}R=\frac{6.17\times {10}^{18}\text{atoms}/{\text{cm}}^{\text{3}}}{4.99\times {10}^{22}\text{atoms}/{\text{cm}}^{\text{3}}}\\ =1.24\times {10}^{-4}\end{array}$

Thus the ratio of arsenic to silicon atoms is $1.24\times {10}^{-4}$.