91Ó°ÊÓ

1. Mass of silicon to be doped, m = 1.0 g
2. Density of silicon,$\mathrm{d}=2330\mathrm{kg}/{\mathrm{m}}^3$
3. Molar mass of silicon,${\mathrm{A}}_{\mathrm{sillicon}}=28.09\mathrm{g}/\mathrm{mol}$
4. Molar mass of phosphorus,${\mathrm{A}}_{\mathrm{phosphorus}}=30.98\mathrm{g}/\mathrm{mol}$
5. Before doping, the density of conduction electrons in the silicon,${\mathrm{n}}_0={10}^{16}{\mathrm{m}}^{-3}$

After doping, the density of conduction electrons in the silicon,$\mathrm{n}{\text{'}}_0={10}^6{\mathrm{n}}_{\mathrm{o}}{\mathrm{m}}^{-3}$

Each phosphorus atom contributes one conduction electron. Thus, doping of silicon atoms requires the replacement of phosphorus atoms with silicon atoms. So, first, we calculate the number of phosphorus atoms in terms of the density of silicon atoms. Then, we need to calculate the number of silicon atoms in the sample. Thus, using this value and the above value of density of phosphorus atoms in terms of silicon atoms in a fraction, we can get the exact number of phosphorus atoms that get doped in place of silicon atoms. Using this value, we can calculate the mass required by using the mass of each phosphorus atom.

Formulae:

The number of conduction electrons or number of atoms per unit volume of a material,

${\mathbf{n}}_{\mathbf{atom}}\mathbf{=}\frac{\mathbf{d}}{{\mathbf{AN}}_{\mathbf{A}}}\mathbf{,}\mathbf{where}\mathbf{}{\mathbf{N}}_{\mathbf{A}}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{022}\mathbf{×}{\mathbf{10}}^{\mathbf{23}}\mathbf{}\mathbf{/}\mathbf{mol}$ (i)

The number of atoms of a material,$\mathbf{N}\mathbf{=}\frac{\mathbf{Given}\mathbf{}\mathbf{mass}}{\mathbf{Mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{one}\mathbf{}\mathbf{atom}\mathbf{}\mathbf{of}\mathbf{}\mathbf{the}\mathbf{}\mathbf{matereial}}$ (ii)

The mass of an atom of a material, $\mathbf{M}\mathbf{=}\mathbf{A}\mathbf{/}{\mathbf{N}}_{\mathbf{A}}$ (iii)

The required mass of the material, m = NM (iv)

Let, be the amount of conduction electrons of phosphorus atoms to be replace the silicon atoms. Thus, we can write that number of phosphorus atoms per silicon atom as follows:

$$\begin{gathered} {\mathrm{n}}_{\mathrm{p}}+{\mathrm{n}}_0={\mathrm{n}}_0\text{'} \\ {\mathrm{n}}_{\mathrm{p}}+{10}^6{\mathrm{n}}_0-{\mathrm{n}}_0\left(âˆ\mu {\mathrm{n}}_0\text{'}={10}^6{\mathrm{n}}_0\right) \\ ≈{10}^6{\mathrm{n}}_0 \\ ={10}^6\left({10}^{16}{\mathrm{m}}^{-3}\right) \\ ={10}^{22}{\mathrm{m}}^{-3} \end{gathered}$$

Now, the number of silicon atoms can be calculated using equation (i) as follows:

$$\begin{gathered} {\mathrm{n}}_{\mathrm{atom}}=\frac{\left(2330\mathrm{kg}/{\mathrm{m}}^3\right)}{\left(0.0281\mathrm{kg}/\mathrm{mol}\right)\left(6.022×{10}^{23}/\mathrm{mol}\right)} \\ =5×10/{\mathrm{m}}^3 \end{gathered}$$

Now, the value of phosphorus atoms to replace one silicon atom can be given by the fractional value as follows:

$$\begin{gathered} \mathrm{frac}=\frac{{10}^{22}{\mathrm{m}}^{-3}}{5×{10}^{28}/{\mathrm{m}}^3} \\ =\frac{1}{5×{10}^6} \end{gathered}$$

Now, the mass of a silicon atom can be given using equation (iii) as follows:

$$\begin{gathered} {\mathrm{m}}_{0,\mathrm{si}}=\frac{28.086\mathrm{g}/\mathrm{mol}}{\left(6.022×{10}^{23}/\mathrm{mol}\right)} \\ =4.66×{10}^{-23}\mathrm{g} \end{gathered}$$

Thus, the number of the silicon atoms in the mass is given using equation (ii) as follows:

$$\begin{gathered} {\mathrm{N}}_{\mathrm{Si}}=\frac{1.0\mathrm{g}}{4.66×{10}^{-23}\mathrm{g}} \\ =2.14×{10}^{22} \end{gathered}$$

Thus, the number of phosphorus atom to dope in1.0 g of silicon can be given as follows:

$$\begin{gathered} {\mathrm{N}}_{\mathrm{p}}=2.14×{10}^{22}/5×{10}^6 \\ =4.29×{10}^{15} \end{gathered}$$

The required mass of the phosphorus is given by substituting equation (iii) in equation (iv) with the given and calculated data as follows:

$$\begin{gathered} \mathrm{m}=\mathrm{N}\left(\mathrm{A}/{\mathrm{N}}_{\mathrm{A}}\right) \\ =\left(4.29×{10}^{15}\right)\left(\frac{30.9758\mathrm{g}/\mathrm{mol}}{6.022×{10}^{23}/\mathrm{mol}}\right) \\ =2.2×{10}^{-7}\mathrm{g} \end{gathered}$$

Hence, the mass of phosphorous is $2.2×{10}^{-7}\mathrm{g}$.