Chapter 29.1, Problem 16PP

To determine

The number of arsenic atoms in per germanium atom.

Answer to Problem 16PP

The number of arsenic atoms in per germanium atom is $1.68\times {10}^{29}$ .

Explanation of Solution

Given data:

Number of atoms per mole in germanium, $N_a=6.02\times {10}^{23}\text{ atom/mole}$

Atomic mass of germanium, $M=72.6$ g

Density of germanium, $\rho =5.23{\text{ g/cm}}^3$

Number of free electrons per atom in germanium, $f=1.6\times {10}^{10}$

Number of free electrons per centimeter cube of arsenic doped in germanium, $e=2\times {10}^3$

Number of free electrons per centimeter cube in germanium, $e\text{'}=2.25\times {10}^{13}$

Formula used:

Free electron per atom in germanium,

  $\frac{\text{Free electron}}{\text{atom}}=\left(\frac{1}{N_a}\right)\left(M\right)\left(\frac{1}{\rho }\right)\left(f\right)$

Free electrons per centimeter cube doped in germanium,

  $\frac{\text{Free electron}}{\text{atom}}=\left(e\right)\left(e\text{'}\right)$

Calculation:

Free electron per atom in germanium,

  $\begin{array}{l}\frac{\text{Free electron}}{\text{atom}}=\left(\frac{1}{N_a}\right)\left(M\right)\left(\frac{1}{\rho }\right)\left(f\right)\\ \frac{\text{Free electron}}{\text{atom}}=\left(\frac{1}{6.02\times {10}^{23}}\right)\left(72.6\right)\left(\frac{1}{5.23}\right)\left(1.16\times {10}^{10}\right)=2.67\times {10}^{-13}\end{array}$

Free electrons per centimeter cube doped in germanium,

  $\begin{array}{l}\frac{\text{Free electron}}{\text{atom}}=\left(e\right)\left(e\text{'}\right)\\ \frac{\text{Free electron}}{\text{atom}}=\left(2\times {10}^3\right)\left(2.25\times {10}^{13}\right)=4.5\times {10}^{16}\\ \frac{\text{Arsenic atoms}}{\text{Germanium atoms}}=\frac{4.5\times {10}^{16}}{2.67\times {10}^{-13}}=1.68\times {10}^{29}\end{array}$

Conclusion:

Hence, the number of arsenic atoms in per germanium atom is $1.68\times {10}^{29}$ .