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# Estimate the Proportion of Boron Impurity Which Will Increase the Conductivity of a Pure Silicon Sample by a Factor of 100. - Physics

Short Note

Estimate the proportion of boron impurity which will increase the conductivity of a pure silicon sample by a factor of 100. Assume that each boron atom creates a hole and the concentration of holes in pure silicon at the same temperature is 7 × 1015 holes per cubic metre. Density of silicon 5 × 1028 atoms per cubic metre.

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#### Solution

Initially, the total number of charge carriers per cubic metre is given by
ni = 2 × 7 × 1015

$\Rightarrow$ ni = 14 × 1015

Finally, the total number of charge carriers per cubic metre is given by
nf = 14 × 1017/m3

We know that the product of the concentrations of holes and conduction electrons remains almost the same.
Let x be the number of holes.
Thus,

$(7 \times {10}^{15} ) \times (7 \times {10}^{15} ) = x \times (14 \times {10}^{17} - x)$

$\Rightarrow 14x \times {10}^{17} - x^2 = 49 \times {10}^{30}$

$\Rightarrow x^2 - 14x \times {10}^{17} - 49 \times {10}^{30} = 0$

$\Rightarrow x = \frac{14 \times {10}^{17} \pm \sqrt{(14 )^2 \times {10}^{34} + 4 \times 49 \times {10}^{30}}}{2}$

$\Rightarrow x = \frac{14 \times {10}^{17} \pm \sqrt{(14 )^2 \times {10}^{34} + 4 \times 49 \times {10}^{30}}}{2}$

$\Rightarrow x = \frac{28 . 0007}{2} \times {10}^{17} = 14 . 00035 \times {10}^{17}$

This is equal to the increased number of holes or the number of atoms of boron added.
Number of atoms of boron added = $(14 . 00035 \times {10}^{17} - 7 \times {10}^{15} ) = 1386 . 035 \times {10}^{15}$

Now, 1386.035 × 1015 atoms are added per 5 × 1028 atoms of Si in 1 m3.
Therefore, 1 atom of boron is added per $\frac{5 \times {10}^{28}}{1386 . 035 \times {10}^{15}}$ atoms of Si in 1 m3.
Proportion of boron impurity is $3 . 607 \times {10}^{- 3} \times {10}^{13} = 3 . 607 \times {10}^{10}$

Concept: Energy Bands in Conductors, Semiconductors and Insulators
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#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 23 Semiconductors and Semiconductor Devices
Q 12 | Page 419
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