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Home » The number of silicon atoms per is This is doped simultaneously with atoms per of Arsenic and per atoms of Indium. Calculate the number of electrons and holes.
The number of silicon atoms per \mathrm{m}^{3} is 5 \times 10^{28} . This is doped simultaneously with \mathbf{5} \times 10^{22} atoms per \mathrm{m}^{3} of Arsenic and 5 \times 10^{20} per \mathrm{m}^{3} atoms of Indium. Calculate the number of electrons and holes.
CBSE | Class 12 | NCERT | Physics | Semiconductor Electronics: Materials, Devices and Simple Circuits
The number of silicon atoms per \mathrm{m}^{3} is 5 \times 10^{28} . This is doped simultaneously with \mathbf{5} \times 10^{22} atoms per \mathrm{m}^{3} of Arsenic and 5 \times 10^{20} per \mathrm{m}^{3} atoms of Indium. Calculate the number of electrons and holes.

Number of silicon atoms is given as N=5 \times 10^{28} atoms / \mathrm{m}^{3}

Number of arsenic atoms is given as \mathrm{n}_{\mathrm{AS}}=5 \times 10^{22} atoms / \mathrm{m}^{3}

Number of indium atoms is given as n_{\ln }=5 \times 10^{22} atoms / \mathrm{m}^{3}

n_{i}=1.5 \times 10^{16} electrons / \mathrm{m}^{3}

n_{e}=5 \times 10^{22}-1.5 \times 10^{16}=4.99 \times 10^{22}

n_{h} be the number of holes.

In the thermal equilibrium, n_{e} n_{h}=n_{i}^{2}

On Calculating, we get

n_{h}=4.51 \times 10^{9}

As a result, the material is a n-type semiconductor as n_{e}>n_{n}.

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