QBManager

Semiconductor Electronics Material Devices and Simple Circuits

150996 Solar energy of $200 {Wm}^{- 2}$ is incident on a horizontal surface. If $20 %$ of this energy can be converted into useful electrical energy, the area needed to supply $8 kW$ energy is

1

150 m^{2}

2

200 m^{2}

3

250 m^{2}

4

100 m^{2}

Explanation:

B Given, Power (P) $= 8 kW$
Solar energy received per square meter $= 200 W m^{- 2}$
Efficiency $= 20 %$
$∵$ According to question we get
$8 \times 10^{3} = 20 % (A \times 200)$
$8 \times 10^{3} = \frac{20}{100} \times A \times 200$
$A = \frac{8 \times 10^{3}}{40}$
$A = 200 m^{2}$

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.

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 $2.0 eV$ . The minimum frequency of the radiation that can be absorbed by the material is nearly

1

10 \times 10^{14} Hz

2

5 \times 10^{14} Hz

3

1 \times 10^{14} Hz

4

20 \times 10^{14} Hz

Explanation:

B Given
$E_{g} = 2.0 eV$
$E_{g} = \frac{hc}{λ}$
$λ = \frac{hc}{E_{g}} = \frac{6.63 \times 10^{- 34} \times 3 \times 10^{8}}{2.0 \times 1.6 \times 10^{- 19}}$
$λ = 6.21 \times 10^{- 7} m$
$∵$ The minimum frequency of radiation
$v = \frac{c}{λ} = \frac{3 \times 10^{8}}{6.21 \times 10^{- 7}}$
$v = 5 \times 10^{14} Hz$

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 $1.9 eV$ . The wavelength of the light emitted will be equal to

1

10.4 \times 10^{- 28} m

2

654 nm

3

654 \AA

4

654 \times 10^{- 11} m

Explanation:

B Given,
Energy group $(E_{g}) = 1.9 eV$
Wavelength of light emitted $(λ) = \frac{hc}{E_{g}}$
$λ = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{1.9 \times 1.6 \times 10^{- 19}}$
$= 651 \times 10^{- 9}$
$λ \approx 650 nm$

NEET- (Odisha)- 2019

Semiconductor Electronics Material Devices and Simple Circuits

150996 Solar energy of $200 {Wm}^{- 2}$ is incident on a horizontal surface. If $20 %$ of this energy can be converted into useful electrical energy, the area needed to supply $8 kW$ energy is

1

150 m^{2}

2

200 m^{2}

3

250 m^{2}

4

100 m^{2}

Explanation:

B Given, Power (P) $= 8 kW$
Solar energy received per square meter $= 200 W m^{- 2}$
Efficiency $= 20 %$
$∵$ According to question we get
$8 \times 10^{3} = 20 % (A \times 200)$
$8 \times 10^{3} = \frac{20}{100} \times A \times 200$
$A = \frac{8 \times 10^{3}}{40}$
$A = 200 m^{2}$

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.

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 $2.0 eV$ . The minimum frequency of the radiation that can be absorbed by the material is nearly

1

10 \times 10^{14} Hz

2

5 \times 10^{14} Hz

3

1 \times 10^{14} Hz

4

20 \times 10^{14} Hz

Explanation:

B Given
$E_{g} = 2.0 eV$
$E_{g} = \frac{hc}{λ}$
$λ = \frac{hc}{E_{g}} = \frac{6.63 \times 10^{- 34} \times 3 \times 10^{8}}{2.0 \times 1.6 \times 10^{- 19}}$
$λ = 6.21 \times 10^{- 7} m$
$∵$ The minimum frequency of radiation
$v = \frac{c}{λ} = \frac{3 \times 10^{8}}{6.21 \times 10^{- 7}}$
$v = 5 \times 10^{14} Hz$

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 $1.9 eV$ . The wavelength of the light emitted will be equal to

1

10.4 \times 10^{- 28} m

2

654 nm

3

654 \AA

4

654 \times 10^{- 11} m

Explanation:

B Given,
Energy group $(E_{g}) = 1.9 eV$
Wavelength of light emitted $(λ) = \frac{hc}{E_{g}}$
$λ = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{1.9 \times 1.6 \times 10^{- 19}}$
$= 651 \times 10^{- 9}$
$λ \approx 650 nm$

NEET- (Odisha)- 2019

Semiconductor Electronics Material Devices and Simple Circuits

150996 Solar energy of $200 {Wm}^{- 2}$ is incident on a horizontal surface. If $20 %$ of this energy can be converted into useful electrical energy, the area needed to supply $8 kW$ energy is

1

150 m^{2}

2

200 m^{2}

3

250 m^{2}

4

100 m^{2}

Explanation:

B Given, Power (P) $= 8 kW$
Solar energy received per square meter $= 200 W m^{- 2}$
Efficiency $= 20 %$
$∵$ According to question we get
$8 \times 10^{3} = 20 % (A \times 200)$
$8 \times 10^{3} = \frac{20}{100} \times A \times 200$
$A = \frac{8 \times 10^{3}}{40}$
$A = 200 m^{2}$

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.

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 $2.0 eV$ . The minimum frequency of the radiation that can be absorbed by the material is nearly

1

10 \times 10^{14} Hz

2

5 \times 10^{14} Hz

3

1 \times 10^{14} Hz

4

20 \times 10^{14} Hz

Explanation:

B Given
$E_{g} = 2.0 eV$
$E_{g} = \frac{hc}{λ}$
$λ = \frac{hc}{E_{g}} = \frac{6.63 \times 10^{- 34} \times 3 \times 10^{8}}{2.0 \times 1.6 \times 10^{- 19}}$
$λ = 6.21 \times 10^{- 7} m$
$∵$ The minimum frequency of radiation
$v = \frac{c}{λ} = \frac{3 \times 10^{8}}{6.21 \times 10^{- 7}}$
$v = 5 \times 10^{14} Hz$

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 $1.9 eV$ . The wavelength of the light emitted will be equal to

1

10.4 \times 10^{- 28} m

2

654 nm

3

654 \AA

4

654 \times 10^{- 11} m

Explanation:

B Given,
Energy group $(E_{g}) = 1.9 eV$
Wavelength of light emitted $(λ) = \frac{hc}{E_{g}}$
$λ = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{1.9 \times 1.6 \times 10^{- 19}}$
$= 651 \times 10^{- 9}$
$λ \approx 650 nm$

NEET- (Odisha)- 2019

Semiconductor Electronics Material Devices and Simple Circuits

150996 Solar energy of $200 {Wm}^{- 2}$ is incident on a horizontal surface. If $20 %$ of this energy can be converted into useful electrical energy, the area needed to supply $8 kW$ energy is

1

150 m^{2}

2

200 m^{2}

3

250 m^{2}

4

100 m^{2}

Explanation:

B Given, Power (P) $= 8 kW$
Solar energy received per square meter $= 200 W m^{- 2}$
Efficiency $= 20 %$
$∵$ According to question we get
$8 \times 10^{3} = 20 % (A \times 200)$
$8 \times 10^{3} = \frac{20}{100} \times A \times 200$
$A = \frac{8 \times 10^{3}}{40}$
$A = 200 m^{2}$

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.

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 $2.0 eV$ . The minimum frequency of the radiation that can be absorbed by the material is nearly

1

10 \times 10^{14} Hz

2

5 \times 10^{14} Hz

3

1 \times 10^{14} Hz

4

20 \times 10^{14} Hz

Explanation:

B Given
$E_{g} = 2.0 eV$
$E_{g} = \frac{hc}{λ}$
$λ = \frac{hc}{E_{g}} = \frac{6.63 \times 10^{- 34} \times 3 \times 10^{8}}{2.0 \times 1.6 \times 10^{- 19}}$
$λ = 6.21 \times 10^{- 7} m$
$∵$ The minimum frequency of radiation
$v = \frac{c}{λ} = \frac{3 \times 10^{8}}{6.21 \times 10^{- 7}}$
$v = 5 \times 10^{14} Hz$

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 $1.9 eV$ . The wavelength of the light emitted will be equal to

1

10.4 \times 10^{- 28} m

2

654 nm

3

654 \AA

4

654 \times 10^{- 11} m

Explanation:

B Given,
Energy group $(E_{g}) = 1.9 eV$
Wavelength of light emitted $(λ) = \frac{hc}{E_{g}}$
$λ = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{1.9 \times 1.6 \times 10^{- 19}}$
$= 651 \times 10^{- 9}$
$λ \approx 650 nm$

NEET- (Odisha)- 2019

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: