QBManager

Semiconductor Electronics Material Devices and Simple Circuits

150984 An LED (Light Emitting Diode) is constructed from a p-n junction based on a certain Ga-As-P semiconducting material whose energy gap is $1.9 eV$ . What is the wavelength of the emitted light?

1

650 nm

2

65 \AA

3

800 nm

4

8000 \AA

Explanation:

B The wavelength of emitted light
$λ = \frac{hc}{E_{g}}$
Where, $E_{g} =$ energy gap of semiconductor $= 1.9 eV$
$λ = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{1.9 \times 1.6 \times 10^{- 19}} m$
$λ = 6.5 \times 10^{- 7} m$
$λ = 650 \times 10^{- 9} m$
$λ = 650 nm$

AIIMS-2010

Semiconductor Electronics Material Devices and Simple Circuits

150985 Which one of the following symbols represents a photodiode?
original image

1

(3)

2 (1)

3

(2)

4 (4)

MHT-CET 2020

Semiconductor Electronics Material Devices and Simple Circuits

150978 Given below are two statements : one is labelled as assertion $A$ and the other is labelled as reason $R$
Assertion A : Photodiodes are used in forward bias usually for measuring the light intensity.
Reason R : For a p-n junction diode, at applied voltage $V$ the current in the forward bias is more than the current in the reverse bias for $| V_{Z} | > \pm V$ $\geq | V_{0} |$ where $V_{0}$ is the threshold voltage and $V_{z}$ is the breakdown voltage.
In the light of the above statements, choose the correct answer from the option given below

1

A

is true but

R

is false

2

A

is false but

R

is true

3 Both

A

and

R

are true but

R

is NOT the correct explanation

A

4 Both

A

and

R

are true and

R

is correct explanation

A

Explanation:

B A photodiode is always operated in reverse bias mode from the photo diodes characteristics it is seen clearly that the photo current is almost independent by applied reverse bias voltage. For zero luminance the photo current is almost zero except for small dark current.
For $P - N$ junction in forward bias current is always greater than reverse bias current i.e. $V \geq V_{0}$

JEE Main-25.01.2023

Semiconductor Electronics Material Devices and Simple Circuits

150991 When a monochromatic point source of light is at a distance of $0.2 m$ from a photocell, the cutoff voltage and the saturation current are respectively $V_{0} = 0.6$ volt and $I_{s} = 18.0 mA$ . If the same source is placed $0.6 m$ away from the photocell, then

1 stopping potential

V_{0} = 0.2

volt and saturation current

I_{s} = 18.0 mA

2 stopping potential

V_{0} = 0.6

volt and saturation current

I_{s} = 18.0 mA

3 stopping potential

V_{0} = 0.6

volt and saturation current

I_{s} = 2.0 mA

4 stopping potential

V_{0} = 0.2

volt and saturation current

I_{s} = 2.0 mA

Explanation:

C The cut-off voltage does not depend on distance between cell \& source,
But intensity is inversely proportional to square of its distance.
Given,
$d = 0.6 m$
$cut-off voltage = 0.6 V$
$I_{sat} \propto I, I \propto \frac{1}{r^{2}}$
$I_{{sat}_{2}} = 18 \times {(\frac{0.2}{0.6})}^{2}$
$= 18 \times \frac{1}{9}$
$I_{sat} = 2 mA$

J and K CET- 2003

Semiconductor Electronics Material Devices and Simple Circuits

150994 The p-n photodiodes are fabricated from semiconductors with band gap of $2.8 eV$ and $2.6 eV$ respectively.
Choose the correct statement from the following.

1 Both diodes will detect

460 nm

light

2 First one will detect

460 nm

light

3 Second one will detect

460 nm

light

4 Both of them will not detect

460 nm

light

Explanation:

C Given,
Energy band gap, $E_{g} = 2.8 eV$
$E_{g}^{'} = 2.6 eV$
Where,
Plank's constant $(h) = 6.6 \times 10^{- 34} J$ -s
Speed of light (c) $= 3 \times 10^{8} m / s$
We know that
$E_{g} = \frac{hc}{λ}$
$λ = \frac{hc}{E_{g}}$
So,
$λ_{1} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{2.8 \times 1.6 \times 10^{- 19}} = 4.4 \times 10^{- 7} m$
$λ_{1} = 441 nm$
$λ_{2} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{2.6 \times 1.6 \times 10^{- 19}} = 4.75 \times 10^{- 7} m$
$λ_{2} = 475 nm$ Hence, the second one will detect $460 nm$ light

TS EAMCET (Medical)-02.05.2018

Semiconductor Electronics Material Devices and Simple Circuits

150984 An LED (Light Emitting Diode) is constructed from a p-n junction based on a certain Ga-As-P semiconducting material whose energy gap is $1.9 eV$ . What is the wavelength of the emitted light?

1

650 nm

2

65 \AA

3

800 nm

4

8000 \AA

Explanation:

B The wavelength of emitted light
$λ = \frac{hc}{E_{g}}$
Where, $E_{g} =$ energy gap of semiconductor $= 1.9 eV$
$λ = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{1.9 \times 1.6 \times 10^{- 19}} m$
$λ = 6.5 \times 10^{- 7} m$
$λ = 650 \times 10^{- 9} m$
$λ = 650 nm$

AIIMS-2010

Semiconductor Electronics Material Devices and Simple Circuits

150985 Which one of the following symbols represents a photodiode?
original image

1

(3)

2 (1)

3

(2)

4 (4)

MHT-CET 2020

Semiconductor Electronics Material Devices and Simple Circuits

150978 Given below are two statements : one is labelled as assertion $A$ and the other is labelled as reason $R$
Assertion A : Photodiodes are used in forward bias usually for measuring the light intensity.
Reason R : For a p-n junction diode, at applied voltage $V$ the current in the forward bias is more than the current in the reverse bias for $| V_{Z} | > \pm V$ $\geq | V_{0} |$ where $V_{0}$ is the threshold voltage and $V_{z}$ is the breakdown voltage.
In the light of the above statements, choose the correct answer from the option given below

1

A

is true but

R

is false

2

A

is false but

R

is true

3 Both

A

and

R

are true but

R

is NOT the correct explanation

A

4 Both

A

and

R

are true and

R

is correct explanation

A

Explanation:

B A photodiode is always operated in reverse bias mode from the photo diodes characteristics it is seen clearly that the photo current is almost independent by applied reverse bias voltage. For zero luminance the photo current is almost zero except for small dark current.
For $P - N$ junction in forward bias current is always greater than reverse bias current i.e. $V \geq V_{0}$

JEE Main-25.01.2023

Semiconductor Electronics Material Devices and Simple Circuits

150991 When a monochromatic point source of light is at a distance of $0.2 m$ from a photocell, the cutoff voltage and the saturation current are respectively $V_{0} = 0.6$ volt and $I_{s} = 18.0 mA$ . If the same source is placed $0.6 m$ away from the photocell, then

1 stopping potential

V_{0} = 0.2

volt and saturation current

I_{s} = 18.0 mA

2 stopping potential

V_{0} = 0.6

volt and saturation current

I_{s} = 18.0 mA

3 stopping potential

V_{0} = 0.6

volt and saturation current

I_{s} = 2.0 mA

4 stopping potential

V_{0} = 0.2

volt and saturation current

I_{s} = 2.0 mA

Explanation:

C The cut-off voltage does not depend on distance between cell \& source,
But intensity is inversely proportional to square of its distance.
Given,
$d = 0.6 m$
$cut-off voltage = 0.6 V$
$I_{sat} \propto I, I \propto \frac{1}{r^{2}}$
$I_{{sat}_{2}} = 18 \times {(\frac{0.2}{0.6})}^{2}$
$= 18 \times \frac{1}{9}$
$I_{sat} = 2 mA$

J and K CET- 2003

Semiconductor Electronics Material Devices and Simple Circuits

150994 The p-n photodiodes are fabricated from semiconductors with band gap of $2.8 eV$ and $2.6 eV$ respectively.
Choose the correct statement from the following.

1 Both diodes will detect

460 nm

light

2 First one will detect

460 nm

light

3 Second one will detect

460 nm

light

4 Both of them will not detect

460 nm

light

Explanation:

C Given,
Energy band gap, $E_{g} = 2.8 eV$
$E_{g}^{'} = 2.6 eV$
Where,
Plank's constant $(h) = 6.6 \times 10^{- 34} J$ -s
Speed of light (c) $= 3 \times 10^{8} m / s$
We know that
$E_{g} = \frac{hc}{λ}$
$λ = \frac{hc}{E_{g}}$
So,
$λ_{1} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{2.8 \times 1.6 \times 10^{- 19}} = 4.4 \times 10^{- 7} m$
$λ_{1} = 441 nm$
$λ_{2} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{2.6 \times 1.6 \times 10^{- 19}} = 4.75 \times 10^{- 7} m$
$λ_{2} = 475 nm$ Hence, the second one will detect $460 nm$ light

TS EAMCET (Medical)-02.05.2018

Semiconductor Electronics Material Devices and Simple Circuits

150984 An LED (Light Emitting Diode) is constructed from a p-n junction based on a certain Ga-As-P semiconducting material whose energy gap is $1.9 eV$ . What is the wavelength of the emitted light?

1

650 nm

2

65 \AA

3

800 nm

4

8000 \AA

Explanation:

B The wavelength of emitted light
$λ = \frac{hc}{E_{g}}$
Where, $E_{g} =$ energy gap of semiconductor $= 1.9 eV$
$λ = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{1.9 \times 1.6 \times 10^{- 19}} m$
$λ = 6.5 \times 10^{- 7} m$
$λ = 650 \times 10^{- 9} m$
$λ = 650 nm$

AIIMS-2010

Semiconductor Electronics Material Devices and Simple Circuits

150985 Which one of the following symbols represents a photodiode?
original image

1

(3)

2 (1)

3

(2)

4 (4)

MHT-CET 2020

Semiconductor Electronics Material Devices and Simple Circuits

150978 Given below are two statements : one is labelled as assertion $A$ and the other is labelled as reason $R$
Assertion A : Photodiodes are used in forward bias usually for measuring the light intensity.
Reason R : For a p-n junction diode, at applied voltage $V$ the current in the forward bias is more than the current in the reverse bias for $| V_{Z} | > \pm V$ $\geq | V_{0} |$ where $V_{0}$ is the threshold voltage and $V_{z}$ is the breakdown voltage.
In the light of the above statements, choose the correct answer from the option given below

1

A

is true but

R

is false

2

A

is false but

R

is true

3 Both

A

and

R

are true but

R

is NOT the correct explanation

A

4 Both

A

and

R

are true and

R

is correct explanation

A

Explanation:

B A photodiode is always operated in reverse bias mode from the photo diodes characteristics it is seen clearly that the photo current is almost independent by applied reverse bias voltage. For zero luminance the photo current is almost zero except for small dark current.
For $P - N$ junction in forward bias current is always greater than reverse bias current i.e. $V \geq V_{0}$

JEE Main-25.01.2023

Semiconductor Electronics Material Devices and Simple Circuits

150991 When a monochromatic point source of light is at a distance of $0.2 m$ from a photocell, the cutoff voltage and the saturation current are respectively $V_{0} = 0.6$ volt and $I_{s} = 18.0 mA$ . If the same source is placed $0.6 m$ away from the photocell, then

1 stopping potential

V_{0} = 0.2

volt and saturation current

I_{s} = 18.0 mA

2 stopping potential

V_{0} = 0.6

volt and saturation current

I_{s} = 18.0 mA

3 stopping potential

V_{0} = 0.6

volt and saturation current

I_{s} = 2.0 mA

4 stopping potential

V_{0} = 0.2

volt and saturation current

I_{s} = 2.0 mA

Explanation:

C The cut-off voltage does not depend on distance between cell \& source,
But intensity is inversely proportional to square of its distance.
Given,
$d = 0.6 m$
$cut-off voltage = 0.6 V$
$I_{sat} \propto I, I \propto \frac{1}{r^{2}}$
$I_{{sat}_{2}} = 18 \times {(\frac{0.2}{0.6})}^{2}$
$= 18 \times \frac{1}{9}$
$I_{sat} = 2 mA$

J and K CET- 2003

Semiconductor Electronics Material Devices and Simple Circuits

150994 The p-n photodiodes are fabricated from semiconductors with band gap of $2.8 eV$ and $2.6 eV$ respectively.
Choose the correct statement from the following.

1 Both diodes will detect

460 nm

light

2 First one will detect

460 nm

light

3 Second one will detect

460 nm

light

4 Both of them will not detect

460 nm

light

Explanation:

C Given,
Energy band gap, $E_{g} = 2.8 eV$
$E_{g}^{'} = 2.6 eV$
Where,
Plank's constant $(h) = 6.6 \times 10^{- 34} J$ -s
Speed of light (c) $= 3 \times 10^{8} m / s$
We know that
$E_{g} = \frac{hc}{λ}$
$λ = \frac{hc}{E_{g}}$
So,
$λ_{1} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{2.8 \times 1.6 \times 10^{- 19}} = 4.4 \times 10^{- 7} m$
$λ_{1} = 441 nm$
$λ_{2} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{2.6 \times 1.6 \times 10^{- 19}} = 4.75 \times 10^{- 7} m$
$λ_{2} = 475 nm$ Hence, the second one will detect $460 nm$ light

TS EAMCET (Medical)-02.05.2018

Semiconductor Electronics Material Devices and Simple Circuits

150984 An LED (Light Emitting Diode) is constructed from a p-n junction based on a certain Ga-As-P semiconducting material whose energy gap is $1.9 eV$ . What is the wavelength of the emitted light?

1

650 nm

2

65 \AA

3

800 nm

4

8000 \AA

Explanation:

B The wavelength of emitted light
$λ = \frac{hc}{E_{g}}$
Where, $E_{g} =$ energy gap of semiconductor $= 1.9 eV$
$λ = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{1.9 \times 1.6 \times 10^{- 19}} m$
$λ = 6.5 \times 10^{- 7} m$
$λ = 650 \times 10^{- 9} m$
$λ = 650 nm$

AIIMS-2010

Semiconductor Electronics Material Devices and Simple Circuits

150985 Which one of the following symbols represents a photodiode?
original image

1

(3)

2 (1)

3

(2)

4 (4)

MHT-CET 2020

Semiconductor Electronics Material Devices and Simple Circuits

150978 Given below are two statements : one is labelled as assertion $A$ and the other is labelled as reason $R$
Assertion A : Photodiodes are used in forward bias usually for measuring the light intensity.
Reason R : For a p-n junction diode, at applied voltage $V$ the current in the forward bias is more than the current in the reverse bias for $| V_{Z} | > \pm V$ $\geq | V_{0} |$ where $V_{0}$ is the threshold voltage and $V_{z}$ is the breakdown voltage.
In the light of the above statements, choose the correct answer from the option given below

1

A

is true but

R

is false

2

A

is false but

R

is true

3 Both

A

and

R

are true but

R

is NOT the correct explanation

A

4 Both

A

and

R

are true and

R

is correct explanation

A

Explanation:

B A photodiode is always operated in reverse bias mode from the photo diodes characteristics it is seen clearly that the photo current is almost independent by applied reverse bias voltage. For zero luminance the photo current is almost zero except for small dark current.
For $P - N$ junction in forward bias current is always greater than reverse bias current i.e. $V \geq V_{0}$

JEE Main-25.01.2023

Semiconductor Electronics Material Devices and Simple Circuits

150991 When a monochromatic point source of light is at a distance of $0.2 m$ from a photocell, the cutoff voltage and the saturation current are respectively $V_{0} = 0.6$ volt and $I_{s} = 18.0 mA$ . If the same source is placed $0.6 m$ away from the photocell, then

1 stopping potential

V_{0} = 0.2

volt and saturation current

I_{s} = 18.0 mA

2 stopping potential

V_{0} = 0.6

volt and saturation current

I_{s} = 18.0 mA

3 stopping potential

V_{0} = 0.6

volt and saturation current

I_{s} = 2.0 mA

4 stopping potential

V_{0} = 0.2

volt and saturation current

I_{s} = 2.0 mA

Explanation:

C The cut-off voltage does not depend on distance between cell \& source,
But intensity is inversely proportional to square of its distance.
Given,
$d = 0.6 m$
$cut-off voltage = 0.6 V$
$I_{sat} \propto I, I \propto \frac{1}{r^{2}}$
$I_{{sat}_{2}} = 18 \times {(\frac{0.2}{0.6})}^{2}$
$= 18 \times \frac{1}{9}$
$I_{sat} = 2 mA$

J and K CET- 2003

Semiconductor Electronics Material Devices and Simple Circuits

150994 The p-n photodiodes are fabricated from semiconductors with band gap of $2.8 eV$ and $2.6 eV$ respectively.
Choose the correct statement from the following.

1 Both diodes will detect

460 nm

light

2 First one will detect

460 nm

light

3 Second one will detect

460 nm

light

4 Both of them will not detect

460 nm

light

Explanation:

C Given,
Energy band gap, $E_{g} = 2.8 eV$
$E_{g}^{'} = 2.6 eV$
Where,
Plank's constant $(h) = 6.6 \times 10^{- 34} J$ -s
Speed of light (c) $= 3 \times 10^{8} m / s$
We know that
$E_{g} = \frac{hc}{λ}$
$λ = \frac{hc}{E_{g}}$
So,
$λ_{1} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{2.8 \times 1.6 \times 10^{- 19}} = 4.4 \times 10^{- 7} m$
$λ_{1} = 441 nm$
$λ_{2} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{2.6 \times 1.6 \times 10^{- 19}} = 4.75 \times 10^{- 7} m$
$λ_{2} = 475 nm$ Hence, the second one will detect $460 nm$ light

TS EAMCET (Medical)-02.05.2018

Semiconductor Electronics Material Devices and Simple Circuits

150984 An LED (Light Emitting Diode) is constructed from a p-n junction based on a certain Ga-As-P semiconducting material whose energy gap is $1.9 eV$ . What is the wavelength of the emitted light?

1

650 nm

2

65 \AA

3

800 nm

4

8000 \AA

Explanation:

B The wavelength of emitted light
$λ = \frac{hc}{E_{g}}$
Where, $E_{g} =$ energy gap of semiconductor $= 1.9 eV$
$λ = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{1.9 \times 1.6 \times 10^{- 19}} m$
$λ = 6.5 \times 10^{- 7} m$
$λ = 650 \times 10^{- 9} m$
$λ = 650 nm$

AIIMS-2010

Semiconductor Electronics Material Devices and Simple Circuits

150985 Which one of the following symbols represents a photodiode?
original image

1

(3)

2 (1)

3

(2)

4 (4)

MHT-CET 2020

Semiconductor Electronics Material Devices and Simple Circuits

150978 Given below are two statements : one is labelled as assertion $A$ and the other is labelled as reason $R$
Assertion A : Photodiodes are used in forward bias usually for measuring the light intensity.
Reason R : For a p-n junction diode, at applied voltage $V$ the current in the forward bias is more than the current in the reverse bias for $| V_{Z} | > \pm V$ $\geq | V_{0} |$ where $V_{0}$ is the threshold voltage and $V_{z}$ is the breakdown voltage.
In the light of the above statements, choose the correct answer from the option given below

1

A

is true but

R

is false

2

A

is false but

R

is true

3 Both

A

and

R

are true but

R

is NOT the correct explanation

A

4 Both

A

and

R

are true and

R

is correct explanation

A

Explanation:

B A photodiode is always operated in reverse bias mode from the photo diodes characteristics it is seen clearly that the photo current is almost independent by applied reverse bias voltage. For zero luminance the photo current is almost zero except for small dark current.
For $P - N$ junction in forward bias current is always greater than reverse bias current i.e. $V \geq V_{0}$

JEE Main-25.01.2023

Semiconductor Electronics Material Devices and Simple Circuits

150991 When a monochromatic point source of light is at a distance of $0.2 m$ from a photocell, the cutoff voltage and the saturation current are respectively $V_{0} = 0.6$ volt and $I_{s} = 18.0 mA$ . If the same source is placed $0.6 m$ away from the photocell, then

1 stopping potential

V_{0} = 0.2

volt and saturation current

I_{s} = 18.0 mA

2 stopping potential

V_{0} = 0.6

volt and saturation current

I_{s} = 18.0 mA

3 stopping potential

V_{0} = 0.6

volt and saturation current

I_{s} = 2.0 mA

4 stopping potential

V_{0} = 0.2

volt and saturation current

I_{s} = 2.0 mA

Explanation:

C The cut-off voltage does not depend on distance between cell \& source,
But intensity is inversely proportional to square of its distance.
Given,
$d = 0.6 m$
$cut-off voltage = 0.6 V$
$I_{sat} \propto I, I \propto \frac{1}{r^{2}}$
$I_{{sat}_{2}} = 18 \times {(\frac{0.2}{0.6})}^{2}$
$= 18 \times \frac{1}{9}$
$I_{sat} = 2 mA$

J and K CET- 2003

Semiconductor Electronics Material Devices and Simple Circuits

150994 The p-n photodiodes are fabricated from semiconductors with band gap of $2.8 eV$ and $2.6 eV$ respectively.
Choose the correct statement from the following.

1 Both diodes will detect

460 nm

light

2 First one will detect

460 nm

light

3 Second one will detect

460 nm

light

4 Both of them will not detect

460 nm

light

Explanation:

C Given,
Energy band gap, $E_{g} = 2.8 eV$
$E_{g}^{'} = 2.6 eV$
Where,
Plank's constant $(h) = 6.6 \times 10^{- 34} J$ -s
Speed of light (c) $= 3 \times 10^{8} m / s$
We know that
$E_{g} = \frac{hc}{λ}$
$λ = \frac{hc}{E_{g}}$
So,
$λ_{1} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{2.8 \times 1.6 \times 10^{- 19}} = 4.4 \times 10^{- 7} m$
$λ_{1} = 441 nm$
$λ_{2} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{2.6 \times 1.6 \times 10^{- 19}} = 4.75 \times 10^{- 7} m$
$λ_{2} = 475 nm$ Hence, the second one will detect $460 nm$ light

TS EAMCET (Medical)-02.05.2018

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