Semiconductor Electronics Material Devices and Simple Circuits
150510
For a pure \(\mathrm{Si}\) crystal has \(5 \times 10^{28}\) atom \(\mathrm{m}^{\mathbf{- 3}}\). It is doped by 1 PPM concentration pentavalent As. Calculate the number of electron \(\&\) holes.
1 \(4.5 \times 10^9 \mathrm{~m}^{-3}\)
2 \(5.4 \times 10^9 \mathrm{~m}^{-3}\)
3 \(4.5 \times 10^{-9} \mathrm{~m}^{-3}\)
4 \(5.4 \times 10^{-9} \mathrm{~m}^{-3}\)
Explanation:
B Given that, No. of Si atoms \(=5 \times 10^{28}\) atoms \(\mathrm{m}^{-3}\) 1 parts per million \(=\frac{1}{10^6}\) No. of pentavalents atoms depend in \(\mathrm{Si}\) crystal \(=\frac{5 \times 10^{28}}{10^6}=5 \times 10^{22} \mathrm{~m}^{-3}\) Number of electrons \(\left(\mathrm{n}_{\mathrm{e}}\right)=5 \times 10^{22}\) And, \(\mathrm{n}_{\mathrm{e}} \mathrm{n}_{\mathrm{h}}=\mathrm{n}_{\mathrm{i}}{ }^2\) \(\mathrm{n}_{\mathrm{h}}=\frac{\mathrm{n}_{\mathrm{i}}^2}{\mathrm{n}_{\mathrm{e}}} \quad\left(\because \mathrm{n}_{\mathrm{i}}=1.5 \times 10^{16}\right)\) \(\mathrm{n}_{\mathrm{h}}=\frac{\left(1.5 \times 10^{16}\right)^2}{5 \times 10^{22}}\) \(\mathrm{n}_{\mathrm{h}}=\frac{2.25 \times 10^{32}}{5 \times 10^{22}}\) \(\mathrm{n}_{\mathrm{h}}=0.45 \times 10^{10}\) \(\mathrm{n}_{\mathrm{h}}=4.5 \times 10^9 \mathrm{~m}^{-3}\)
GUJCET 2020
Semiconductor Electronics Material Devices and Simple Circuits
150511
The Forbidden gap between conduction band \& valance band is maximum for
1 Metal
2 Insulator
3 Supercondutor
4 Semiconductor
Explanation:
B According to band theory each material has two energy band with different levels (i) Valence band (ii) Conduction band - The difference between these two energy level is called forbidden energy gap. - Insulator are material, which have the maximum value of forbidden energy gap. Therefore, the band gap for insulator is largest.
GUJCET 2020
Semiconductor Electronics Material Devices and Simple Circuits
150512
For \(P N\) junction, the intensity of electric field is \(1 \times 10^6 \mathrm{~V} / \mathrm{m}\) and the width of depletion region is 5000 . The value of potential barrier \(=\) V.
1 0.05
2 0.005
3 0.5
4 5
Explanation:
C Given, The intensity of electric field \((\mathrm{E})=1 \times 10^6 \mathrm{~V} / \mathrm{m}\) Width of depletion \(=5000 \AA\) We know that, Electric field intensity \((E)=\frac{V}{d}\) \(\therefore \quad \mathrm{V}=\mathrm{E} \times \mathrm{d}\) \(\mathrm{V}=10^6 \times 5 \times 10^{-7}\)So, \(\quad \mathrm{V}=0.5 \mathrm{~V}\)
GUJCET 2019
Semiconductor Electronics Material Devices and Simple Circuits
150519
Assertion: A n-type semiconductor has a large number of electrons but still it is electrically neutral. Reason: A n-type semiconductor is obtained by doping an intrinsic semiconductor with a pentavalent impurity.
1 If both assertion and reason are true and reason is the correct explanation of assertion.
2 If both assertion and reason are true but reason is not the correct explanation of assertion.
3 If assertion is true but reason is false.
4 If both assertion and reason are false.
Explanation:
B A n-type semiconductor is formed by doping pure germanium or silicon crystal with suitable impurity atoms of valency five. Trivalent impurities are added which introduces an extra hole which requires less energy for conduction in p-type semiconductor. In n-type semiconductor as the impurity atoms take the position of Ge atom in germanium crystal, its four electrons form covalent bonds by sharing electrons form covalent bonds by sharing electrons with the neighbouring four atoms of germanium where the \(5^{\text {th }}\) electron is left free. Since the atom on the whole is electrically natural, the n-type semiconductor is also neutral.
Semiconductor Electronics Material Devices and Simple Circuits
150510
For a pure \(\mathrm{Si}\) crystal has \(5 \times 10^{28}\) atom \(\mathrm{m}^{\mathbf{- 3}}\). It is doped by 1 PPM concentration pentavalent As. Calculate the number of electron \(\&\) holes.
1 \(4.5 \times 10^9 \mathrm{~m}^{-3}\)
2 \(5.4 \times 10^9 \mathrm{~m}^{-3}\)
3 \(4.5 \times 10^{-9} \mathrm{~m}^{-3}\)
4 \(5.4 \times 10^{-9} \mathrm{~m}^{-3}\)
Explanation:
B Given that, No. of Si atoms \(=5 \times 10^{28}\) atoms \(\mathrm{m}^{-3}\) 1 parts per million \(=\frac{1}{10^6}\) No. of pentavalents atoms depend in \(\mathrm{Si}\) crystal \(=\frac{5 \times 10^{28}}{10^6}=5 \times 10^{22} \mathrm{~m}^{-3}\) Number of electrons \(\left(\mathrm{n}_{\mathrm{e}}\right)=5 \times 10^{22}\) And, \(\mathrm{n}_{\mathrm{e}} \mathrm{n}_{\mathrm{h}}=\mathrm{n}_{\mathrm{i}}{ }^2\) \(\mathrm{n}_{\mathrm{h}}=\frac{\mathrm{n}_{\mathrm{i}}^2}{\mathrm{n}_{\mathrm{e}}} \quad\left(\because \mathrm{n}_{\mathrm{i}}=1.5 \times 10^{16}\right)\) \(\mathrm{n}_{\mathrm{h}}=\frac{\left(1.5 \times 10^{16}\right)^2}{5 \times 10^{22}}\) \(\mathrm{n}_{\mathrm{h}}=\frac{2.25 \times 10^{32}}{5 \times 10^{22}}\) \(\mathrm{n}_{\mathrm{h}}=0.45 \times 10^{10}\) \(\mathrm{n}_{\mathrm{h}}=4.5 \times 10^9 \mathrm{~m}^{-3}\)
GUJCET 2020
Semiconductor Electronics Material Devices and Simple Circuits
150511
The Forbidden gap between conduction band \& valance band is maximum for
1 Metal
2 Insulator
3 Supercondutor
4 Semiconductor
Explanation:
B According to band theory each material has two energy band with different levels (i) Valence band (ii) Conduction band - The difference between these two energy level is called forbidden energy gap. - Insulator are material, which have the maximum value of forbidden energy gap. Therefore, the band gap for insulator is largest.
GUJCET 2020
Semiconductor Electronics Material Devices and Simple Circuits
150512
For \(P N\) junction, the intensity of electric field is \(1 \times 10^6 \mathrm{~V} / \mathrm{m}\) and the width of depletion region is 5000 . The value of potential barrier \(=\) V.
1 0.05
2 0.005
3 0.5
4 5
Explanation:
C Given, The intensity of electric field \((\mathrm{E})=1 \times 10^6 \mathrm{~V} / \mathrm{m}\) Width of depletion \(=5000 \AA\) We know that, Electric field intensity \((E)=\frac{V}{d}\) \(\therefore \quad \mathrm{V}=\mathrm{E} \times \mathrm{d}\) \(\mathrm{V}=10^6 \times 5 \times 10^{-7}\)So, \(\quad \mathrm{V}=0.5 \mathrm{~V}\)
GUJCET 2019
Semiconductor Electronics Material Devices and Simple Circuits
150519
Assertion: A n-type semiconductor has a large number of electrons but still it is electrically neutral. Reason: A n-type semiconductor is obtained by doping an intrinsic semiconductor with a pentavalent impurity.
1 If both assertion and reason are true and reason is the correct explanation of assertion.
2 If both assertion and reason are true but reason is not the correct explanation of assertion.
3 If assertion is true but reason is false.
4 If both assertion and reason are false.
Explanation:
B A n-type semiconductor is formed by doping pure germanium or silicon crystal with suitable impurity atoms of valency five. Trivalent impurities are added which introduces an extra hole which requires less energy for conduction in p-type semiconductor. In n-type semiconductor as the impurity atoms take the position of Ge atom in germanium crystal, its four electrons form covalent bonds by sharing electrons form covalent bonds by sharing electrons with the neighbouring four atoms of germanium where the \(5^{\text {th }}\) electron is left free. Since the atom on the whole is electrically natural, the n-type semiconductor is also neutral.
Semiconductor Electronics Material Devices and Simple Circuits
150510
For a pure \(\mathrm{Si}\) crystal has \(5 \times 10^{28}\) atom \(\mathrm{m}^{\mathbf{- 3}}\). It is doped by 1 PPM concentration pentavalent As. Calculate the number of electron \(\&\) holes.
1 \(4.5 \times 10^9 \mathrm{~m}^{-3}\)
2 \(5.4 \times 10^9 \mathrm{~m}^{-3}\)
3 \(4.5 \times 10^{-9} \mathrm{~m}^{-3}\)
4 \(5.4 \times 10^{-9} \mathrm{~m}^{-3}\)
Explanation:
B Given that, No. of Si atoms \(=5 \times 10^{28}\) atoms \(\mathrm{m}^{-3}\) 1 parts per million \(=\frac{1}{10^6}\) No. of pentavalents atoms depend in \(\mathrm{Si}\) crystal \(=\frac{5 \times 10^{28}}{10^6}=5 \times 10^{22} \mathrm{~m}^{-3}\) Number of electrons \(\left(\mathrm{n}_{\mathrm{e}}\right)=5 \times 10^{22}\) And, \(\mathrm{n}_{\mathrm{e}} \mathrm{n}_{\mathrm{h}}=\mathrm{n}_{\mathrm{i}}{ }^2\) \(\mathrm{n}_{\mathrm{h}}=\frac{\mathrm{n}_{\mathrm{i}}^2}{\mathrm{n}_{\mathrm{e}}} \quad\left(\because \mathrm{n}_{\mathrm{i}}=1.5 \times 10^{16}\right)\) \(\mathrm{n}_{\mathrm{h}}=\frac{\left(1.5 \times 10^{16}\right)^2}{5 \times 10^{22}}\) \(\mathrm{n}_{\mathrm{h}}=\frac{2.25 \times 10^{32}}{5 \times 10^{22}}\) \(\mathrm{n}_{\mathrm{h}}=0.45 \times 10^{10}\) \(\mathrm{n}_{\mathrm{h}}=4.5 \times 10^9 \mathrm{~m}^{-3}\)
GUJCET 2020
Semiconductor Electronics Material Devices and Simple Circuits
150511
The Forbidden gap between conduction band \& valance band is maximum for
1 Metal
2 Insulator
3 Supercondutor
4 Semiconductor
Explanation:
B According to band theory each material has two energy band with different levels (i) Valence band (ii) Conduction band - The difference between these two energy level is called forbidden energy gap. - Insulator are material, which have the maximum value of forbidden energy gap. Therefore, the band gap for insulator is largest.
GUJCET 2020
Semiconductor Electronics Material Devices and Simple Circuits
150512
For \(P N\) junction, the intensity of electric field is \(1 \times 10^6 \mathrm{~V} / \mathrm{m}\) and the width of depletion region is 5000 . The value of potential barrier \(=\) V.
1 0.05
2 0.005
3 0.5
4 5
Explanation:
C Given, The intensity of electric field \((\mathrm{E})=1 \times 10^6 \mathrm{~V} / \mathrm{m}\) Width of depletion \(=5000 \AA\) We know that, Electric field intensity \((E)=\frac{V}{d}\) \(\therefore \quad \mathrm{V}=\mathrm{E} \times \mathrm{d}\) \(\mathrm{V}=10^6 \times 5 \times 10^{-7}\)So, \(\quad \mathrm{V}=0.5 \mathrm{~V}\)
GUJCET 2019
Semiconductor Electronics Material Devices and Simple Circuits
150519
Assertion: A n-type semiconductor has a large number of electrons but still it is electrically neutral. Reason: A n-type semiconductor is obtained by doping an intrinsic semiconductor with a pentavalent impurity.
1 If both assertion and reason are true and reason is the correct explanation of assertion.
2 If both assertion and reason are true but reason is not the correct explanation of assertion.
3 If assertion is true but reason is false.
4 If both assertion and reason are false.
Explanation:
B A n-type semiconductor is formed by doping pure germanium or silicon crystal with suitable impurity atoms of valency five. Trivalent impurities are added which introduces an extra hole which requires less energy for conduction in p-type semiconductor. In n-type semiconductor as the impurity atoms take the position of Ge atom in germanium crystal, its four electrons form covalent bonds by sharing electrons form covalent bonds by sharing electrons with the neighbouring four atoms of germanium where the \(5^{\text {th }}\) electron is left free. Since the atom on the whole is electrically natural, the n-type semiconductor is also neutral.
NEET Test Series from KOTA - 10 Papers In MS WORD
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Semiconductor Electronics Material Devices and Simple Circuits
150510
For a pure \(\mathrm{Si}\) crystal has \(5 \times 10^{28}\) atom \(\mathrm{m}^{\mathbf{- 3}}\). It is doped by 1 PPM concentration pentavalent As. Calculate the number of electron \(\&\) holes.
1 \(4.5 \times 10^9 \mathrm{~m}^{-3}\)
2 \(5.4 \times 10^9 \mathrm{~m}^{-3}\)
3 \(4.5 \times 10^{-9} \mathrm{~m}^{-3}\)
4 \(5.4 \times 10^{-9} \mathrm{~m}^{-3}\)
Explanation:
B Given that, No. of Si atoms \(=5 \times 10^{28}\) atoms \(\mathrm{m}^{-3}\) 1 parts per million \(=\frac{1}{10^6}\) No. of pentavalents atoms depend in \(\mathrm{Si}\) crystal \(=\frac{5 \times 10^{28}}{10^6}=5 \times 10^{22} \mathrm{~m}^{-3}\) Number of electrons \(\left(\mathrm{n}_{\mathrm{e}}\right)=5 \times 10^{22}\) And, \(\mathrm{n}_{\mathrm{e}} \mathrm{n}_{\mathrm{h}}=\mathrm{n}_{\mathrm{i}}{ }^2\) \(\mathrm{n}_{\mathrm{h}}=\frac{\mathrm{n}_{\mathrm{i}}^2}{\mathrm{n}_{\mathrm{e}}} \quad\left(\because \mathrm{n}_{\mathrm{i}}=1.5 \times 10^{16}\right)\) \(\mathrm{n}_{\mathrm{h}}=\frac{\left(1.5 \times 10^{16}\right)^2}{5 \times 10^{22}}\) \(\mathrm{n}_{\mathrm{h}}=\frac{2.25 \times 10^{32}}{5 \times 10^{22}}\) \(\mathrm{n}_{\mathrm{h}}=0.45 \times 10^{10}\) \(\mathrm{n}_{\mathrm{h}}=4.5 \times 10^9 \mathrm{~m}^{-3}\)
GUJCET 2020
Semiconductor Electronics Material Devices and Simple Circuits
150511
The Forbidden gap between conduction band \& valance band is maximum for
1 Metal
2 Insulator
3 Supercondutor
4 Semiconductor
Explanation:
B According to band theory each material has two energy band with different levels (i) Valence band (ii) Conduction band - The difference between these two energy level is called forbidden energy gap. - Insulator are material, which have the maximum value of forbidden energy gap. Therefore, the band gap for insulator is largest.
GUJCET 2020
Semiconductor Electronics Material Devices and Simple Circuits
150512
For \(P N\) junction, the intensity of electric field is \(1 \times 10^6 \mathrm{~V} / \mathrm{m}\) and the width of depletion region is 5000 . The value of potential barrier \(=\) V.
1 0.05
2 0.005
3 0.5
4 5
Explanation:
C Given, The intensity of electric field \((\mathrm{E})=1 \times 10^6 \mathrm{~V} / \mathrm{m}\) Width of depletion \(=5000 \AA\) We know that, Electric field intensity \((E)=\frac{V}{d}\) \(\therefore \quad \mathrm{V}=\mathrm{E} \times \mathrm{d}\) \(\mathrm{V}=10^6 \times 5 \times 10^{-7}\)So, \(\quad \mathrm{V}=0.5 \mathrm{~V}\)
GUJCET 2019
Semiconductor Electronics Material Devices and Simple Circuits
150519
Assertion: A n-type semiconductor has a large number of electrons but still it is electrically neutral. Reason: A n-type semiconductor is obtained by doping an intrinsic semiconductor with a pentavalent impurity.
1 If both assertion and reason are true and reason is the correct explanation of assertion.
2 If both assertion and reason are true but reason is not the correct explanation of assertion.
3 If assertion is true but reason is false.
4 If both assertion and reason are false.
Explanation:
B A n-type semiconductor is formed by doping pure germanium or silicon crystal with suitable impurity atoms of valency five. Trivalent impurities are added which introduces an extra hole which requires less energy for conduction in p-type semiconductor. In n-type semiconductor as the impurity atoms take the position of Ge atom in germanium crystal, its four electrons form covalent bonds by sharing electrons form covalent bonds by sharing electrons with the neighbouring four atoms of germanium where the \(5^{\text {th }}\) electron is left free. Since the atom on the whole is electrically natural, the n-type semiconductor is also neutral.
