CBSE Class 12 · 2024 · SemiconductorsMultiple Choice

A pure Si crystal having

5 \times 1 0^{28}

atoms

m^{- 3}

is doped with 1 ppm concentration of antimony. If the concentration of holes in the doped crystal is found to be

4.5 \times 1 0^{9} m^{- 3}

, the concentration (in

m^{- 3}

) of intrinsic charge carriers in the Si crystal is about

Answer: (A)

Solution

Antimony is pentavalent, contributing one donor electron per atom.

n_{e} = \frac{5 \times 1 0 ^{28}}{1 0 ^{6}} = 5 \times 1 0^{22} m^{- 3} .

By the mass-action law

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

n_{i} = (5 \times 1 0^{22}) (4.5 \times 1 0^{9}) = 2.25 \times 1 0^{32} = 1.5 \times 1 0^{16} m^{- 3} .

n_{i} \approx 1.5 \times 1 0^{16} m^{- 3}

CBSE Class 12 Board · 2024 · Single correct option