
In an intrinsic semiconductor, the density of conduction electrons is $7.07 \times {10^{15}}{m^{ - 3}}$ .When it is doped with indium, the density of holes becomes $5 \times {10^{22}}{m^{ - 3}}$ . Find the density of conduction electrons in doped semiconductor.
A. $0$
B. $1 \times {10^9}{m^{ - 3}}$
C. $7 \times {10^{15}}{m^{ - 3}}$
D. $5 \times {10^{22}}{m^{ - 3}}$
Answer
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Hint: Intrinsic semiconductor is a type of semiconductor which is in pure state and when it is doped with some impurities, it changes into extrinsic semiconductor. In a doped semiconductor, ${n_i}^2 = {n_e}{n_h}$ where ${n_e}$ and ${n_h}$ are the no. density of electrons and holes respectively and ${n_i}$is the no. density of charge carries i.e. electrons or holes in the intrinsic semiconductor.
Complete step by step answer:
We should know that the semiconductors are basically of two types Intrinsic and Extrinsic. Intrinsic semiconductor is a type of semiconductor which is in pure state and when it is doped with some impurities, it changes into extrinsic semiconductor. So, according to the question, indium is that impurity which doped in intrinsic semiconductor.
As in a doped semiconductor, ${n_i}^2 = {n_e}{n_h}$ where ${n_e}$ and ${n_h}$ are the no. density of electrons and holes respectively and ${n_i}$is the no. density of charge carries i.e. electrons or holes in the intrinsic semiconductor.
So, according to the question the no. density of electrons of intrinsic semiconductor is given ${n_i} = 7.07 \times {10^{15}}{m^{ - 3}}$ and no. density of holes when it is doped is given ${n_h} = 5 \times {10^{22}}{m^{ - 3}}$ .
So, substituting these values in the equation we have
${\left( {7.07 \times {{10}^{15}}} \right)^2} = {n_i} \times \left( {5 \times {{10}^{22}}} \right)$
On solving this we have the no. density of conduction electrons as
${n_i} = 0.99 \times {10^9} \approx 1 \times {10^9}{m^{ - 3}}$
Hence, option B is correct.
Note: The doping in the intrinsic semiconductor is done to improve its functionality and this is done in two different ways. When it is doped with the elements like As, Sb, Bi etc. in which electrons is in majority is known as n-type semiconductor whereas when is doped with the elements like Al, B, In etc. in which holes comes in majority is known as p-type semiconductor.
Complete step by step answer:
We should know that the semiconductors are basically of two types Intrinsic and Extrinsic. Intrinsic semiconductor is a type of semiconductor which is in pure state and when it is doped with some impurities, it changes into extrinsic semiconductor. So, according to the question, indium is that impurity which doped in intrinsic semiconductor.
As in a doped semiconductor, ${n_i}^2 = {n_e}{n_h}$ where ${n_e}$ and ${n_h}$ are the no. density of electrons and holes respectively and ${n_i}$is the no. density of charge carries i.e. electrons or holes in the intrinsic semiconductor.
So, according to the question the no. density of electrons of intrinsic semiconductor is given ${n_i} = 7.07 \times {10^{15}}{m^{ - 3}}$ and no. density of holes when it is doped is given ${n_h} = 5 \times {10^{22}}{m^{ - 3}}$ .
So, substituting these values in the equation we have
${\left( {7.07 \times {{10}^{15}}} \right)^2} = {n_i} \times \left( {5 \times {{10}^{22}}} \right)$
On solving this we have the no. density of conduction electrons as
${n_i} = 0.99 \times {10^9} \approx 1 \times {10^9}{m^{ - 3}}$
Hence, option B is correct.
Note: The doping in the intrinsic semiconductor is done to improve its functionality and this is done in two different ways. When it is doped with the elements like As, Sb, Bi etc. in which electrons is in majority is known as n-type semiconductor whereas when is doped with the elements like Al, B, In etc. in which holes comes in majority is known as p-type semiconductor.
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