Chapter 29, Problem 21P

To determine

The factor by which the density of conduction electrons increased.

Answer to Problem 21P

  $5\times {10}^6$ .

Explanation of Solution

Given:

Density ${\rho }_{si}=2330kg/m^3$

Density of conduction electrons, ${10}^{16}electrons/m^3$

Formula used:

It is known that: ${\text{N}}_{\text{Si}}\text{=}\frac{\text{a mole of pure silicon}}{{\text{ρ}}_{\text{Si}}}\times \text{density of conduction electrons}$

Where,

  ${\text{N}}_{\text{Si}}$ is the number of conduction electrons.

  ${\text{ρ}}_{\text{Si}}$ is the density of silicon.

Calculation:

Let a mole of pure silicon is $28g$ .

Substitute the data in the formula:

  ${\text{N}}_{\text{Si}}\text{=}\frac{\text{a mole of pure silicon}}{{\text{ρ}}_{\text{Si}}}\times \text{density of conduction electrons}$

  ${\text{N}}_{\text{Si}}\text{=}\left[\frac{28g}{2330{\text{kg/m}}^{\text{3}}}\right]\times {10}^{16}{\text{electrons/m}}^{\text{3}}=1.20\times {10}^{11}$

The more conduction electrons is provided by the doping:

  ${\text{N}}_{\text{doping}}\text{=}\left[\frac{6.02\times {10}^{23}\text{atoms}}{{10}^6}\right]=6.02\times {10}^{17}$

Therefore, the density of conduction electrons is increases:

  $\frac{N_{doping}}{N_{Si}}=\frac{6.02\times {10}^{17}}{1.20\times {10}^{11}}=5\times {10}^6$

Conclusion: The density is increased by $5\times {10}^6$ .