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PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365597 If an intrinsic semiconductor is heated, the ratio of free electrons to holes is

1 Greater than one

2 Less than one

3 Equal to one

4 Dercease and becomes zero

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365598 Due to diffusion of electrons from $n$ to $p$ -side.
I.
An ionised acceptor is left in the $p$ -region
II.
An ionised donor is left in the $n$ -region
III.
Electrons of $n$ -side comes to $p$ -side and electron-hole combination takes place in $p$ -side

1 I and II

2 II and III

3 I and III

4 I, II and III

Explanation:

When an electron diffuses from $n \to p$ , it leaves behind an ionised donor (species which has become ion by donating electron) on $n$ -side.
This ionised donor (positive charge) is immobile as it is bonded to the surrounding atoms. As the electrons continue to diffuse from $n$ - $p$ , a layer of positive charge (or positive space-charge region) on $n$ -side of the junction is developed. On $p$ -side atom receiving electrons are ionised acceptor. So, option (4) is correct.

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365599 The difference in the variation of resistance with temperature in a metal and a semiconductor arises essentially due to the difference in the :

1 Crystal structure

2 Variation of the number of charge carriers with temperature

3 Type of bonding

4 Variation of scattering mechanism with temperature

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365600 In an intrinsic semiconductor, at an ordinary temperature, the correct relation between the number of electrons per unit volume $n_{e}$ and number of holes per unit volume $n_{e}$

1

n_{e} = n_{h}

2

n_{e} > n_{h}

3

n_{e} < n_{h}

4

n_{e} = n_{h} = 0

Explanation:

In intrinsic semiconductor at ordinary temperature, number of free electrons per unit volume $(n_{e}) =$ number of holes per unit volume $(n_{h})$

MHTCET - 2020

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365597 If an intrinsic semiconductor is heated, the ratio of free electrons to holes is

1 Greater than one

2 Less than one

3 Equal to one

4 Dercease and becomes zero

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365598 Due to diffusion of electrons from $n$ to $p$ -side.
I.
An ionised acceptor is left in the $p$ -region
II.
An ionised donor is left in the $n$ -region
III.
Electrons of $n$ -side comes to $p$ -side and electron-hole combination takes place in $p$ -side

1 I and II

2 II and III

3 I and III

4 I, II and III

Explanation:

When an electron diffuses from $n \to p$ , it leaves behind an ionised donor (species which has become ion by donating electron) on $n$ -side.
This ionised donor (positive charge) is immobile as it is bonded to the surrounding atoms. As the electrons continue to diffuse from $n$ - $p$ , a layer of positive charge (or positive space-charge region) on $n$ -side of the junction is developed. On $p$ -side atom receiving electrons are ionised acceptor. So, option (4) is correct.

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365599 The difference in the variation of resistance with temperature in a metal and a semiconductor arises essentially due to the difference in the :

1 Crystal structure

2 Variation of the number of charge carriers with temperature

3 Type of bonding

4 Variation of scattering mechanism with temperature

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365600 In an intrinsic semiconductor, at an ordinary temperature, the correct relation between the number of electrons per unit volume $n_{e}$ and number of holes per unit volume $n_{e}$

1

n_{e} = n_{h}

2

n_{e} > n_{h}

3

n_{e} < n_{h}

4

n_{e} = n_{h} = 0

Explanation:

In intrinsic semiconductor at ordinary temperature, number of free electrons per unit volume $(n_{e}) =$ number of holes per unit volume $(n_{h})$

MHTCET - 2020

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365597 If an intrinsic semiconductor is heated, the ratio of free electrons to holes is

1 Greater than one

2 Less than one

3 Equal to one

4 Dercease and becomes zero

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365598 Due to diffusion of electrons from $n$ to $p$ -side.
I.
An ionised acceptor is left in the $p$ -region
II.
An ionised donor is left in the $n$ -region
III.
Electrons of $n$ -side comes to $p$ -side and electron-hole combination takes place in $p$ -side

1 I and II

2 II and III

3 I and III

4 I, II and III

Explanation:

When an electron diffuses from $n \to p$ , it leaves behind an ionised donor (species which has become ion by donating electron) on $n$ -side.
This ionised donor (positive charge) is immobile as it is bonded to the surrounding atoms. As the electrons continue to diffuse from $n$ - $p$ , a layer of positive charge (or positive space-charge region) on $n$ -side of the junction is developed. On $p$ -side atom receiving electrons are ionised acceptor. So, option (4) is correct.

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365599 The difference in the variation of resistance with temperature in a metal and a semiconductor arises essentially due to the difference in the :

1 Crystal structure

2 Variation of the number of charge carriers with temperature

3 Type of bonding

4 Variation of scattering mechanism with temperature

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365600 In an intrinsic semiconductor, at an ordinary temperature, the correct relation between the number of electrons per unit volume $n_{e}$ and number of holes per unit volume $n_{e}$

1

n_{e} = n_{h}

2

n_{e} > n_{h}

3

n_{e} < n_{h}

4

n_{e} = n_{h} = 0

Explanation:

In intrinsic semiconductor at ordinary temperature, number of free electrons per unit volume $(n_{e}) =$ number of holes per unit volume $(n_{h})$

MHTCET - 2020

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365597 If an intrinsic semiconductor is heated, the ratio of free electrons to holes is

1 Greater than one

2 Less than one

3 Equal to one

4 Dercease and becomes zero

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365598 Due to diffusion of electrons from $n$ to $p$ -side.
I.
An ionised acceptor is left in the $p$ -region
II.
An ionised donor is left in the $n$ -region
III.
Electrons of $n$ -side comes to $p$ -side and electron-hole combination takes place in $p$ -side

1 I and II

2 II and III

3 I and III

4 I, II and III

Explanation:

When an electron diffuses from $n \to p$ , it leaves behind an ionised donor (species which has become ion by donating electron) on $n$ -side.
This ionised donor (positive charge) is immobile as it is bonded to the surrounding atoms. As the electrons continue to diffuse from $n$ - $p$ , a layer of positive charge (or positive space-charge region) on $n$ -side of the junction is developed. On $p$ -side atom receiving electrons are ionised acceptor. So, option (4) is correct.

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365599 The difference in the variation of resistance with temperature in a metal and a semiconductor arises essentially due to the difference in the :

1 Crystal structure

2 Variation of the number of charge carriers with temperature

3 Type of bonding

4 Variation of scattering mechanism with temperature

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365600 In an intrinsic semiconductor, at an ordinary temperature, the correct relation between the number of electrons per unit volume $n_{e}$ and number of holes per unit volume $n_{e}$

1

n_{e} = n_{h}

2

n_{e} > n_{h}

3

n_{e} < n_{h}

4

n_{e} = n_{h} = 0

Explanation:

In intrinsic semiconductor at ordinary temperature, number of free electrons per unit volume $(n_{e}) =$ number of holes per unit volume $(n_{h})$

MHTCET - 2020

Question Bank Series

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