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Question: A pure Si crystal has \(5 \times 10^{22}\) atoms \(m^{- 3}\) it is doped by 1 ppm concentration of p...

A pure Si crystal has 5×10225 \times 10^{22} atoms m−3m^{- 3} it is doped by 1 ppm concentration of pentavalent As. The number of holes is [n2i=npne]\left\lbrack {n^{2}}_{i} = n_{p}n_{e} \right\rbrack

(Take ni=1.5×1016m−3n_{i} = 1.5 \times 10^{16}m^{- 3})

A

4.5×109m−34.5 \times 10^{9}m^{- 3}

B

4.5×106m−34.5 \times 10^{6}m^{- 3}

C

2.5×109m−32.5 \times 10^{9}m^{- 3}

D

2.5×106m−32.5 \times 10^{6}m^{- 3}

Answer

4.5×109m−34.5 \times 10^{9}m^{- 3}

Explanation

Solution

: The electron and holes concentration in a semiconductor in thermal equilibrium is

nenh=ni2n_{e}n_{h} = n_{i}^{2}

Or nh=ni2nen_{h} = \frac{n_{i}^{2}}{n_{e}}

Here, ni=1.5×1016m−3,ne=5×1022m−3n_{i} = 1.5 \times 10^{16}m^{- 3},n_{e} = 5 \times 10^{22}m^{- 3}

∴nh=(1.5×1016)2(5×1022)=2.25×10325×1022=4.5×109m−3\therefore n_{h} = \frac{(1.5 \times 10^{16})^{2}}{(5 \times 10^{22})} = \frac{2.25 \times 10^{32}}{5 \times 10^{22}} = 4.5 \times 10^{9}m^{- 3}