Determine the proportion of boron impurity required to increase the...

The conductivity of pure (intrinsic) silicon is given by:

$\sigma_i = n_i e \mu_e + n_i e \mu_h$

Assuming the mobilities of electrons and holes are approximately equal ($\mu_e \approx \mu_h = \mu$), we get:

$\sigma_i = 2 n_i e \mu$

When boron (a trivalent impurity) is added, the silicon becomes a p-type semiconductor. The conductivity is predominantly due to holes:

$\sigma_p \approx n_h e \mu_h = N_A e \mu$

where $N_A$ is the concentration of boron atoms.

Given that the conductivity increases by a factor of $100$:

$\sigma_p = 100 \sigma_i$

$N_A e \mu = 100 (2 n_i e \mu)$

$N_A = 200 n_i$

Substituting the given intrinsic hole concentration $n_i = 7 \times 10^{15} \text{ m}^{-3}$:

$N_A = 200 \times 7 \times 10^{15} = 1.4 \times 10^{18} \text{ atoms/m}^3$

The density of silicon atoms is $N_{Si} = 5 \times 10^{28} \text{ atoms/m}^3$. The proportion of boron impurity required is:

$\text{Proportion} = \dfrac{N_A}{N_{Si}} = \dfrac{1.4 \times 10^{18}}{5 \times 10^{28}} = 2.8 \times 10^{-11}$

To express this as $1$ part in $x$:

$x = \dfrac{1}{2.8 \times 10^{-11}} = \dfrac{10^{11}}{2.8} \approx 3.57 \times 10^{10}$

Thus, the required proportion is $1$ in $3.57 \times 10^{10}$.

Answer: $1$ in $3.57 \times 10^{10}$