Question

What mass of phosphorous is needed to dope $1.0\text{g}$ of silicon so that the number density of conduction electrons in the silicon increased by a multiply factor of ${10}^6$ from the ${10}^{16}{\text{m}}^{-3}$ in pure silicon.
(Take the concentration of silicon atoms $n_{\text{Si}}=5\times {10}^{28}{\text{m}}^{-3}\text{Molar mass of silicon}{M}_{\text{Si}}=28.1\text{g}\text{Molar mass of phosphorous}{M}_{\text{P}}=30.97\text{g}\text{Avogadro number}{N}_A=6.023\times {10}^{23})$

• A. $1.5\times {10}^{-7}\text{g}$
• B. $2\times {10}^{-7}\text{g}$
• C. $2.20\times {10}^{-7}\text{g}$
• D. $4.4\times {10}^{-9}\text{g}$

Answer 1

The correct option is C $2.20\times {10}^{-7}\text{g}$

Each phosporous atom contributes one conduction electron.

Let $n_0$ be the number of conduction electrons in pure silicon at room temperature which equals

$n_0={10}^{16}{\text{m}}^{-3}$,

we want to increase this number by a factor of a million $\left({10}^6\right)$,

so we can write :

${10}^6{n}_0=n_0+n_p$

Where $n_p$ is the number density of the phosphorous atoms,

$\Rightarrow ({10}^6-1)n_0=n_p$

$\Rightarrow {10}^6{n}_0=n_p[\because {10}^6-1\approx {10}^6]$

$\Rightarrow n_p={10}^{22}{\text{m}}^{-3}$

So, we need to add ${10}^{22}$ atoms of phosphorous per cubic meter.

Given that, the concentration of silicon atoms,

$n_{\text{Si}}=5.0\times {10}^{28}{m}^{-3}$

$\Rightarrow \frac{n_{\text{Si}}}{n_{\text{p}}}=\frac{5.0\times {10}^{28}{\text{m}}^{-3}}{{10}^{22}{\text{m}}^{-3}}=5\times {10}^6$

This equation tell us that, If we replace only one silicon atom in five million, with a phosphorous atom, the number of electrons in the conduction band will be increased by a factor of a ${10}^6$.

Now, the number of silicon atoms in $1\text{g}$ is given by,

$N_{\text{Si}}=\frac{m_{\text{Si}}{N}_A}{M_{\text{Si}}}$

Substituting the values, we get,

$N_{\text{Si}}=\frac{\left(1.0\text{g}\right)(6.023\times {10}^{23}{\text{mol}}^{-1})}{28.1{\text{g mol}}^{-1}}$

$\Rightarrow N_{\text{Si}}=2.143\times {10}^{22}$

Since one of every $5\times {10}^6$ silicon atoms is replaced with a phosphorous atom, then we can write :

$N_{\text{P}}=\frac{N_{\text{Si}}}{5\times {10}^6}=\frac{2.143\times {10}^{22}}{5\times {10}^6}$

$\Rightarrow N_{\text{P}}=4.28\times {10}^{15}$

Similarity, the mass of phosphorous containing $N_P$ number of atoms is given by,

$m_p=\frac{N_P{M}_P}{N_A}$

$m_P=\frac{(4.286\times {10}^{15})\left(30.97{\text{g mol}}^{-1}\right)}{(6.023\times {10}^{23}{\text{mol}}^{-1})}$

$m_P=2.20\times {10}^{-7}\text{g}$

Hence, option $\left(C\right)$ is correct.