Semiconductor Electronics Material Devices and Simple Circuits
150996
Solar energy of \(200 \mathrm{Wm}^{-2}\) is incident on a horizontal surface. If \(20 \%\) of this energy can be converted into useful electrical energy, the area needed to supply \(8 \mathrm{~kW}\) energy is
1 \(150 \mathrm{~m}^2\)
2 \(200 \mathrm{~m}^2\)
3 \(250 \mathrm{~m}^2\)
4 \(100 \mathrm{~m}^2\)
Explanation:
B Given, Power (P) \(=8 \mathrm{~kW}\) Solar energy received per square meter \(=200 \mathrm{~W} \mathrm{~m}^{-2}\) Efficiency \(=20 \%\) \(\because\) According to question we get \(8 \times 10^3=20 \%(\mathrm{~A} \times 200)\) \(8 \times 10^3=\frac{20}{100} \times \mathrm{A} \times 200\) \(\mathrm{~A}=\frac{8 \times 10^3}{40}\) \(\mathrm{~A}=200 \mathrm{~m}^2\)
Assam CEE-2018
Semiconductor Electronics Material Devices and Simple Circuits
151004
Assertion: Photodiode and solar cell work on same mechanism. Reason: Area is large for solar cell.
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 Photodiodes and solar cell are type of semiconducting device working on same principal or mechanism. Hence assertion is true. We know that solar panel have a large area and therefore makes it possible for the solar cells to emit enough electron to generate electricity but in photodiode does not like that, So reason is true in case of photodiode. Thus, Assertion and reason is true but reason is not correct explanation for assertion.
AIIMS-26.05.2019(M) Shift-1
Semiconductor Electronics Material Devices and Simple Circuits
151007
A p-n photodiode is made of a material with a band gap of \(2.0 \mathrm{eV}\). The minimum frequency of the radiation that can be absorbed by the material is nearly
1 \(10 \times 10^{14} \mathrm{~Hz}\)
2 \(5 \times 10^{14} \mathrm{~Hz}\)
3 \(1 \times 10^{14} \mathrm{~Hz}\)
4 \(20 \times 10^{14} \mathrm{~Hz}\)
Explanation:
B Given \(\mathrm{E}_{\mathrm{g}}=2.0 \mathrm{eV}\) \(\mathrm{E}_{\mathrm{g}}=\frac{\mathrm{hc}}{\lambda}\) \(\lambda=\frac{\mathrm{hc}}{\mathrm{E}_{\mathrm{g}}}=\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{2.0 \times 1.6 \times 10^{-19}}\) \(\lambda=6.21 \times 10^{-7} \mathrm{~m}\) \(\because\) The minimum frequency of radiation \(v=\frac{\mathrm{c}}{\lambda}=\frac{3 \times 10^8}{6.21 \times 10^{-7}}\) \(v=5 \times 10^{14} \mathrm{~Hz}\)
MHT-CET 2008
Semiconductor Electronics Material Devices and Simple Circuits
151013
An LED is constructed from a p-n junction diode using GaAsP. The energy gap is \(1.9 \mathrm{eV}\). The wavelength of the light emitted will be equal to
1 \(10.4 \times 10^{-28} \mathrm{~m}\)
2 \(654 \mathrm{~nm}\)
3 \(654 \AA\)
4 \(654 \times 10^{-11} \mathrm{~m}\)
Explanation:
B Given, Energy group \(\left(\mathrm{E}_{\mathrm{g}}\right)=1.9 \mathrm{eV}\) Wavelength of light emitted \((\lambda)=\frac{\mathrm{hc}}{\mathrm{E}_{\mathrm{g}}}\) \(\lambda =\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{1.9 \times 1.6 \times 10^{-19}}\) \(=651 \times 10^{-9}\) \(\lambda \approx 650 \mathrm{~nm}\)
Semiconductor Electronics Material Devices and Simple Circuits
150996
Solar energy of \(200 \mathrm{Wm}^{-2}\) is incident on a horizontal surface. If \(20 \%\) of this energy can be converted into useful electrical energy, the area needed to supply \(8 \mathrm{~kW}\) energy is
1 \(150 \mathrm{~m}^2\)
2 \(200 \mathrm{~m}^2\)
3 \(250 \mathrm{~m}^2\)
4 \(100 \mathrm{~m}^2\)
Explanation:
B Given, Power (P) \(=8 \mathrm{~kW}\) Solar energy received per square meter \(=200 \mathrm{~W} \mathrm{~m}^{-2}\) Efficiency \(=20 \%\) \(\because\) According to question we get \(8 \times 10^3=20 \%(\mathrm{~A} \times 200)\) \(8 \times 10^3=\frac{20}{100} \times \mathrm{A} \times 200\) \(\mathrm{~A}=\frac{8 \times 10^3}{40}\) \(\mathrm{~A}=200 \mathrm{~m}^2\)
Assam CEE-2018
Semiconductor Electronics Material Devices and Simple Circuits
151004
Assertion: Photodiode and solar cell work on same mechanism. Reason: Area is large for solar cell.
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 Photodiodes and solar cell are type of semiconducting device working on same principal or mechanism. Hence assertion is true. We know that solar panel have a large area and therefore makes it possible for the solar cells to emit enough electron to generate electricity but in photodiode does not like that, So reason is true in case of photodiode. Thus, Assertion and reason is true but reason is not correct explanation for assertion.
AIIMS-26.05.2019(M) Shift-1
Semiconductor Electronics Material Devices and Simple Circuits
151007
A p-n photodiode is made of a material with a band gap of \(2.0 \mathrm{eV}\). The minimum frequency of the radiation that can be absorbed by the material is nearly
1 \(10 \times 10^{14} \mathrm{~Hz}\)
2 \(5 \times 10^{14} \mathrm{~Hz}\)
3 \(1 \times 10^{14} \mathrm{~Hz}\)
4 \(20 \times 10^{14} \mathrm{~Hz}\)
Explanation:
B Given \(\mathrm{E}_{\mathrm{g}}=2.0 \mathrm{eV}\) \(\mathrm{E}_{\mathrm{g}}=\frac{\mathrm{hc}}{\lambda}\) \(\lambda=\frac{\mathrm{hc}}{\mathrm{E}_{\mathrm{g}}}=\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{2.0 \times 1.6 \times 10^{-19}}\) \(\lambda=6.21 \times 10^{-7} \mathrm{~m}\) \(\because\) The minimum frequency of radiation \(v=\frac{\mathrm{c}}{\lambda}=\frac{3 \times 10^8}{6.21 \times 10^{-7}}\) \(v=5 \times 10^{14} \mathrm{~Hz}\)
MHT-CET 2008
Semiconductor Electronics Material Devices and Simple Circuits
151013
An LED is constructed from a p-n junction diode using GaAsP. The energy gap is \(1.9 \mathrm{eV}\). The wavelength of the light emitted will be equal to
1 \(10.4 \times 10^{-28} \mathrm{~m}\)
2 \(654 \mathrm{~nm}\)
3 \(654 \AA\)
4 \(654 \times 10^{-11} \mathrm{~m}\)
Explanation:
B Given, Energy group \(\left(\mathrm{E}_{\mathrm{g}}\right)=1.9 \mathrm{eV}\) Wavelength of light emitted \((\lambda)=\frac{\mathrm{hc}}{\mathrm{E}_{\mathrm{g}}}\) \(\lambda =\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{1.9 \times 1.6 \times 10^{-19}}\) \(=651 \times 10^{-9}\) \(\lambda \approx 650 \mathrm{~nm}\)
Semiconductor Electronics Material Devices and Simple Circuits
150996
Solar energy of \(200 \mathrm{Wm}^{-2}\) is incident on a horizontal surface. If \(20 \%\) of this energy can be converted into useful electrical energy, the area needed to supply \(8 \mathrm{~kW}\) energy is
1 \(150 \mathrm{~m}^2\)
2 \(200 \mathrm{~m}^2\)
3 \(250 \mathrm{~m}^2\)
4 \(100 \mathrm{~m}^2\)
Explanation:
B Given, Power (P) \(=8 \mathrm{~kW}\) Solar energy received per square meter \(=200 \mathrm{~W} \mathrm{~m}^{-2}\) Efficiency \(=20 \%\) \(\because\) According to question we get \(8 \times 10^3=20 \%(\mathrm{~A} \times 200)\) \(8 \times 10^3=\frac{20}{100} \times \mathrm{A} \times 200\) \(\mathrm{~A}=\frac{8 \times 10^3}{40}\) \(\mathrm{~A}=200 \mathrm{~m}^2\)
Assam CEE-2018
Semiconductor Electronics Material Devices and Simple Circuits
151004
Assertion: Photodiode and solar cell work on same mechanism. Reason: Area is large for solar cell.
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 Photodiodes and solar cell are type of semiconducting device working on same principal or mechanism. Hence assertion is true. We know that solar panel have a large area and therefore makes it possible for the solar cells to emit enough electron to generate electricity but in photodiode does not like that, So reason is true in case of photodiode. Thus, Assertion and reason is true but reason is not correct explanation for assertion.
AIIMS-26.05.2019(M) Shift-1
Semiconductor Electronics Material Devices and Simple Circuits
151007
A p-n photodiode is made of a material with a band gap of \(2.0 \mathrm{eV}\). The minimum frequency of the radiation that can be absorbed by the material is nearly
1 \(10 \times 10^{14} \mathrm{~Hz}\)
2 \(5 \times 10^{14} \mathrm{~Hz}\)
3 \(1 \times 10^{14} \mathrm{~Hz}\)
4 \(20 \times 10^{14} \mathrm{~Hz}\)
Explanation:
B Given \(\mathrm{E}_{\mathrm{g}}=2.0 \mathrm{eV}\) \(\mathrm{E}_{\mathrm{g}}=\frac{\mathrm{hc}}{\lambda}\) \(\lambda=\frac{\mathrm{hc}}{\mathrm{E}_{\mathrm{g}}}=\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{2.0 \times 1.6 \times 10^{-19}}\) \(\lambda=6.21 \times 10^{-7} \mathrm{~m}\) \(\because\) The minimum frequency of radiation \(v=\frac{\mathrm{c}}{\lambda}=\frac{3 \times 10^8}{6.21 \times 10^{-7}}\) \(v=5 \times 10^{14} \mathrm{~Hz}\)
MHT-CET 2008
Semiconductor Electronics Material Devices and Simple Circuits
151013
An LED is constructed from a p-n junction diode using GaAsP. The energy gap is \(1.9 \mathrm{eV}\). The wavelength of the light emitted will be equal to
1 \(10.4 \times 10^{-28} \mathrm{~m}\)
2 \(654 \mathrm{~nm}\)
3 \(654 \AA\)
4 \(654 \times 10^{-11} \mathrm{~m}\)
Explanation:
B Given, Energy group \(\left(\mathrm{E}_{\mathrm{g}}\right)=1.9 \mathrm{eV}\) Wavelength of light emitted \((\lambda)=\frac{\mathrm{hc}}{\mathrm{E}_{\mathrm{g}}}\) \(\lambda =\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{1.9 \times 1.6 \times 10^{-19}}\) \(=651 \times 10^{-9}\) \(\lambda \approx 650 \mathrm{~nm}\)
Semiconductor Electronics Material Devices and Simple Circuits
150996
Solar energy of \(200 \mathrm{Wm}^{-2}\) is incident on a horizontal surface. If \(20 \%\) of this energy can be converted into useful electrical energy, the area needed to supply \(8 \mathrm{~kW}\) energy is
1 \(150 \mathrm{~m}^2\)
2 \(200 \mathrm{~m}^2\)
3 \(250 \mathrm{~m}^2\)
4 \(100 \mathrm{~m}^2\)
Explanation:
B Given, Power (P) \(=8 \mathrm{~kW}\) Solar energy received per square meter \(=200 \mathrm{~W} \mathrm{~m}^{-2}\) Efficiency \(=20 \%\) \(\because\) According to question we get \(8 \times 10^3=20 \%(\mathrm{~A} \times 200)\) \(8 \times 10^3=\frac{20}{100} \times \mathrm{A} \times 200\) \(\mathrm{~A}=\frac{8 \times 10^3}{40}\) \(\mathrm{~A}=200 \mathrm{~m}^2\)
Assam CEE-2018
Semiconductor Electronics Material Devices and Simple Circuits
151004
Assertion: Photodiode and solar cell work on same mechanism. Reason: Area is large for solar cell.
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 Photodiodes and solar cell are type of semiconducting device working on same principal or mechanism. Hence assertion is true. We know that solar panel have a large area and therefore makes it possible for the solar cells to emit enough electron to generate electricity but in photodiode does not like that, So reason is true in case of photodiode. Thus, Assertion and reason is true but reason is not correct explanation for assertion.
AIIMS-26.05.2019(M) Shift-1
Semiconductor Electronics Material Devices and Simple Circuits
151007
A p-n photodiode is made of a material with a band gap of \(2.0 \mathrm{eV}\). The minimum frequency of the radiation that can be absorbed by the material is nearly
1 \(10 \times 10^{14} \mathrm{~Hz}\)
2 \(5 \times 10^{14} \mathrm{~Hz}\)
3 \(1 \times 10^{14} \mathrm{~Hz}\)
4 \(20 \times 10^{14} \mathrm{~Hz}\)
Explanation:
B Given \(\mathrm{E}_{\mathrm{g}}=2.0 \mathrm{eV}\) \(\mathrm{E}_{\mathrm{g}}=\frac{\mathrm{hc}}{\lambda}\) \(\lambda=\frac{\mathrm{hc}}{\mathrm{E}_{\mathrm{g}}}=\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{2.0 \times 1.6 \times 10^{-19}}\) \(\lambda=6.21 \times 10^{-7} \mathrm{~m}\) \(\because\) The minimum frequency of radiation \(v=\frac{\mathrm{c}}{\lambda}=\frac{3 \times 10^8}{6.21 \times 10^{-7}}\) \(v=5 \times 10^{14} \mathrm{~Hz}\)
MHT-CET 2008
Semiconductor Electronics Material Devices and Simple Circuits
151013
An LED is constructed from a p-n junction diode using GaAsP. The energy gap is \(1.9 \mathrm{eV}\). The wavelength of the light emitted will be equal to
1 \(10.4 \times 10^{-28} \mathrm{~m}\)
2 \(654 \mathrm{~nm}\)
3 \(654 \AA\)
4 \(654 \times 10^{-11} \mathrm{~m}\)
Explanation:
B Given, Energy group \(\left(\mathrm{E}_{\mathrm{g}}\right)=1.9 \mathrm{eV}\) Wavelength of light emitted \((\lambda)=\frac{\mathrm{hc}}{\mathrm{E}_{\mathrm{g}}}\) \(\lambda =\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{1.9 \times 1.6 \times 10^{-19}}\) \(=651 \times 10^{-9}\) \(\lambda \approx 650 \mathrm{~nm}\)
