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Semiconductor Electronics Material Devices and Simple Circuits

150549 A p-type semiconductor has acceptor levels 57 me $V$ above the valence band. The maximum wavelength of light required to create a hole

1

57 \AA

2

57 \times 10^{- 3} \AA

3

217105 \AA

4

11.61 \times 10^{- 33} \AA

Explanation:

C Given,
Energy $(E) = 57 meV$
We know that,
$E = \frac{hc}{λ}$
$∴ λ = \frac{hc}{E}$
$λ = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{57 \times 10^{- 3} \times 1.6 \times 10^{- 19}}$
$λ = 217105 \AA$

Manipal UGET-2011

Semiconductor Electronics Material Devices and Simple Circuits

150556 Mobilities of electrons and holes in a sample of intrinsic germanium at room temperature are $0.36 m^{2} V^{- 1} s^{- 1}$ and $0.17 m^{2} V^{- 1} s^{- 1}$ . The electron and hole densities are each equal to $2.5 \times$ $10^{19} m^{3}$ . The electrical conductivity of germanium is

1

4.24 {Sm}^{- 1}

2

2.12 {Sm}^{- 1}

3

1.09 {Sm}^{- 1}

4

0.47 {Sm}^{- 1}

Explanation:

B Given that,
$μ_{e} = 0.36 m^{2} V^{- 1} s^{- 1}$
$μ_{H} = 0.17 m^{2} V^{- 1} s^{- 1}$
Then electrical conductivity of germanium,
$σ = \frac{1}{ρ} = e (μ_{e} n_{e} + μ_{n} n_{n})$
$σ = 1.6 \times 10^{- 19} (0.36 \times 2.5 \times 10^{19} + 0.17 \times 2.5 \times 10^{19})$
$σ = 2.12 second m^{- 1}$

VITEEE-2013

Semiconductor Electronics Material Devices and Simple Circuits

150557 A potential difference of $2 V$ is applied between the opposite faces of a Ge crystal plate of area 1 ${cm}^{2}$ and thickness $0.5 mm$ . If the concentration of electrons in Ge is $2 \times 10^{19} / m^{2}$ and mobilities of electrons and holes are $0.36 m^{2} V^{- 1} s^{- 1}$ and 0.14 $m^{2} V^{- 1} s^{- 1}$ respectively, then the current flowing through the plate will be

1

0.25 A

2

0.45 A

3

0.56 A

4

0.64 A

Explanation:

D Given,
$V = 2 volt$
$l = 0.5 mm = 0.5 \times 10^{- 3} m$
$A = 1 {cm}^{2} = 10^{- 4} m^{2}$
$n = 2 \times 10^{19} / m^{2}$
$μ_{e} = 0.36 m^{2} V^{- 1} s^{- 1}$
$μ_{h} = 0.14 m^{2} V^{- 1} s^{- 1}$
We know conductivity-
$σ = ne (μ_{c} + μ_{h})$
$σ = 2 \times 10^{19} \times 1.6 \times 10^{- 19} (0.36 + 0.14)$
$σ = 1.6 (Ω m)^{- 1}$
Then,
$R = ρ \frac{l}{A} = \frac{l}{σ A} = \frac{0.5 \times 10^{- 3}}{1.6 \times 10^{- 4}} = \frac{25}{8} Ω$ So, $i = \frac{V}{R} = \frac{2}{25 / 8} = 0.64 A$

VITEEE-2010

Semiconductor Electronics Material Devices and Simple Circuits

150559 A conductor and a semi-conductor are connected in parallel as shown in the figure. At a certain voltage both ammeters registers the same current. If the voltage of the DC source is increased then
original image

1 the ammeter connected to the semiconductor will register higher current than the ammeter connected to the conductor

2 the ammeter connected to the conductor will register higher current than the ammeter connected to the semiconductor

3 the ammeters connected to both semiconductor and conductor will register the same current

4 the ammeter connected to both semiconductor and conductor will register no change in the current

Explanation:

C As per given figure, here, conductor and semiconductor are connected in parallel. Therefore, voltage across them will be the same. If ammeters show the same reading then their resistances $R = \frac{V}{I}$ will be the same. If voltage is increased by small value then following the same relation $V \propto I$ for constant $R$ .
Hence, both conductor and semiconductor will display the same currents.

VITEEE-2007

Semiconductor Electronics Material Devices and Simple Circuits

150549 A p-type semiconductor has acceptor levels 57 me $V$ above the valence band. The maximum wavelength of light required to create a hole

1

57 \AA

2

57 \times 10^{- 3} \AA

3

217105 \AA

4

11.61 \times 10^{- 33} \AA

Explanation:

C Given,
Energy $(E) = 57 meV$
We know that,
$E = \frac{hc}{λ}$
$∴ λ = \frac{hc}{E}$
$λ = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{57 \times 10^{- 3} \times 1.6 \times 10^{- 19}}$
$λ = 217105 \AA$

Manipal UGET-2011

Semiconductor Electronics Material Devices and Simple Circuits

150556 Mobilities of electrons and holes in a sample of intrinsic germanium at room temperature are $0.36 m^{2} V^{- 1} s^{- 1}$ and $0.17 m^{2} V^{- 1} s^{- 1}$ . The electron and hole densities are each equal to $2.5 \times$ $10^{19} m^{3}$ . The electrical conductivity of germanium is

1

4.24 {Sm}^{- 1}

2

2.12 {Sm}^{- 1}

3

1.09 {Sm}^{- 1}

4

0.47 {Sm}^{- 1}

Explanation:

B Given that,
$μ_{e} = 0.36 m^{2} V^{- 1} s^{- 1}$
$μ_{H} = 0.17 m^{2} V^{- 1} s^{- 1}$
Then electrical conductivity of germanium,
$σ = \frac{1}{ρ} = e (μ_{e} n_{e} + μ_{n} n_{n})$
$σ = 1.6 \times 10^{- 19} (0.36 \times 2.5 \times 10^{19} + 0.17 \times 2.5 \times 10^{19})$
$σ = 2.12 second m^{- 1}$

VITEEE-2013

Semiconductor Electronics Material Devices and Simple Circuits

150557 A potential difference of $2 V$ is applied between the opposite faces of a Ge crystal plate of area 1 ${cm}^{2}$ and thickness $0.5 mm$ . If the concentration of electrons in Ge is $2 \times 10^{19} / m^{2}$ and mobilities of electrons and holes are $0.36 m^{2} V^{- 1} s^{- 1}$ and 0.14 $m^{2} V^{- 1} s^{- 1}$ respectively, then the current flowing through the plate will be

1

0.25 A

2

0.45 A

3

0.56 A

4

0.64 A

Explanation:

D Given,
$V = 2 volt$
$l = 0.5 mm = 0.5 \times 10^{- 3} m$
$A = 1 {cm}^{2} = 10^{- 4} m^{2}$
$n = 2 \times 10^{19} / m^{2}$
$μ_{e} = 0.36 m^{2} V^{- 1} s^{- 1}$
$μ_{h} = 0.14 m^{2} V^{- 1} s^{- 1}$
We know conductivity-
$σ = ne (μ_{c} + μ_{h})$
$σ = 2 \times 10^{19} \times 1.6 \times 10^{- 19} (0.36 + 0.14)$
$σ = 1.6 (Ω m)^{- 1}$
Then,
$R = ρ \frac{l}{A} = \frac{l}{σ A} = \frac{0.5 \times 10^{- 3}}{1.6 \times 10^{- 4}} = \frac{25}{8} Ω$ So, $i = \frac{V}{R} = \frac{2}{25 / 8} = 0.64 A$

VITEEE-2010

Semiconductor Electronics Material Devices and Simple Circuits

150559 A conductor and a semi-conductor are connected in parallel as shown in the figure. At a certain voltage both ammeters registers the same current. If the voltage of the DC source is increased then
original image

1 the ammeter connected to the semiconductor will register higher current than the ammeter connected to the conductor

2 the ammeter connected to the conductor will register higher current than the ammeter connected to the semiconductor

3 the ammeters connected to both semiconductor and conductor will register the same current

4 the ammeter connected to both semiconductor and conductor will register no change in the current

Explanation:

C As per given figure, here, conductor and semiconductor are connected in parallel. Therefore, voltage across them will be the same. If ammeters show the same reading then their resistances $R = \frac{V}{I}$ will be the same. If voltage is increased by small value then following the same relation $V \propto I$ for constant $R$ .
Hence, both conductor and semiconductor will display the same currents.

VITEEE-2007

Semiconductor Electronics Material Devices and Simple Circuits

150549 A p-type semiconductor has acceptor levels 57 me $V$ above the valence band. The maximum wavelength of light required to create a hole

1

57 \AA

2

57 \times 10^{- 3} \AA

3

217105 \AA

4

11.61 \times 10^{- 33} \AA

Explanation:

C Given,
Energy $(E) = 57 meV$
We know that,
$E = \frac{hc}{λ}$
$∴ λ = \frac{hc}{E}$
$λ = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{57 \times 10^{- 3} \times 1.6 \times 10^{- 19}}$
$λ = 217105 \AA$

Manipal UGET-2011

Semiconductor Electronics Material Devices and Simple Circuits

150556 Mobilities of electrons and holes in a sample of intrinsic germanium at room temperature are $0.36 m^{2} V^{- 1} s^{- 1}$ and $0.17 m^{2} V^{- 1} s^{- 1}$ . The electron and hole densities are each equal to $2.5 \times$ $10^{19} m^{3}$ . The electrical conductivity of germanium is

1

4.24 {Sm}^{- 1}

2

2.12 {Sm}^{- 1}

3

1.09 {Sm}^{- 1}

4

0.47 {Sm}^{- 1}

Explanation:

B Given that,
$μ_{e} = 0.36 m^{2} V^{- 1} s^{- 1}$
$μ_{H} = 0.17 m^{2} V^{- 1} s^{- 1}$
Then electrical conductivity of germanium,
$σ = \frac{1}{ρ} = e (μ_{e} n_{e} + μ_{n} n_{n})$
$σ = 1.6 \times 10^{- 19} (0.36 \times 2.5 \times 10^{19} + 0.17 \times 2.5 \times 10^{19})$
$σ = 2.12 second m^{- 1}$

VITEEE-2013

Semiconductor Electronics Material Devices and Simple Circuits

150557 A potential difference of $2 V$ is applied between the opposite faces of a Ge crystal plate of area 1 ${cm}^{2}$ and thickness $0.5 mm$ . If the concentration of electrons in Ge is $2 \times 10^{19} / m^{2}$ and mobilities of electrons and holes are $0.36 m^{2} V^{- 1} s^{- 1}$ and 0.14 $m^{2} V^{- 1} s^{- 1}$ respectively, then the current flowing through the plate will be

1

0.25 A

2

0.45 A

3

0.56 A

4

0.64 A

Explanation:

D Given,
$V = 2 volt$
$l = 0.5 mm = 0.5 \times 10^{- 3} m$
$A = 1 {cm}^{2} = 10^{- 4} m^{2}$
$n = 2 \times 10^{19} / m^{2}$
$μ_{e} = 0.36 m^{2} V^{- 1} s^{- 1}$
$μ_{h} = 0.14 m^{2} V^{- 1} s^{- 1}$
We know conductivity-
$σ = ne (μ_{c} + μ_{h})$
$σ = 2 \times 10^{19} \times 1.6 \times 10^{- 19} (0.36 + 0.14)$
$σ = 1.6 (Ω m)^{- 1}$
Then,
$R = ρ \frac{l}{A} = \frac{l}{σ A} = \frac{0.5 \times 10^{- 3}}{1.6 \times 10^{- 4}} = \frac{25}{8} Ω$ So, $i = \frac{V}{R} = \frac{2}{25 / 8} = 0.64 A$

VITEEE-2010

Semiconductor Electronics Material Devices and Simple Circuits

150559 A conductor and a semi-conductor are connected in parallel as shown in the figure. At a certain voltage both ammeters registers the same current. If the voltage of the DC source is increased then
original image

1 the ammeter connected to the semiconductor will register higher current than the ammeter connected to the conductor

2 the ammeter connected to the conductor will register higher current than the ammeter connected to the semiconductor

3 the ammeters connected to both semiconductor and conductor will register the same current

4 the ammeter connected to both semiconductor and conductor will register no change in the current

Explanation:

C As per given figure, here, conductor and semiconductor are connected in parallel. Therefore, voltage across them will be the same. If ammeters show the same reading then their resistances $R = \frac{V}{I}$ will be the same. If voltage is increased by small value then following the same relation $V \propto I$ for constant $R$ .
Hence, both conductor and semiconductor will display the same currents.

VITEEE-2007

Semiconductor Electronics Material Devices and Simple Circuits

150549 A p-type semiconductor has acceptor levels 57 me $V$ above the valence band. The maximum wavelength of light required to create a hole

1

57 \AA

2

57 \times 10^{- 3} \AA

3

217105 \AA

4

11.61 \times 10^{- 33} \AA

Explanation:

C Given,
Energy $(E) = 57 meV$
We know that,
$E = \frac{hc}{λ}$
$∴ λ = \frac{hc}{E}$
$λ = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{57 \times 10^{- 3} \times 1.6 \times 10^{- 19}}$
$λ = 217105 \AA$

Manipal UGET-2011

Semiconductor Electronics Material Devices and Simple Circuits

150556 Mobilities of electrons and holes in a sample of intrinsic germanium at room temperature are $0.36 m^{2} V^{- 1} s^{- 1}$ and $0.17 m^{2} V^{- 1} s^{- 1}$ . The electron and hole densities are each equal to $2.5 \times$ $10^{19} m^{3}$ . The electrical conductivity of germanium is

1

4.24 {Sm}^{- 1}

2

2.12 {Sm}^{- 1}

3

1.09 {Sm}^{- 1}

4

0.47 {Sm}^{- 1}

Explanation:

B Given that,
$μ_{e} = 0.36 m^{2} V^{- 1} s^{- 1}$
$μ_{H} = 0.17 m^{2} V^{- 1} s^{- 1}$
Then electrical conductivity of germanium,
$σ = \frac{1}{ρ} = e (μ_{e} n_{e} + μ_{n} n_{n})$
$σ = 1.6 \times 10^{- 19} (0.36 \times 2.5 \times 10^{19} + 0.17 \times 2.5 \times 10^{19})$
$σ = 2.12 second m^{- 1}$

VITEEE-2013

Semiconductor Electronics Material Devices and Simple Circuits

150557 A potential difference of $2 V$ is applied between the opposite faces of a Ge crystal plate of area 1 ${cm}^{2}$ and thickness $0.5 mm$ . If the concentration of electrons in Ge is $2 \times 10^{19} / m^{2}$ and mobilities of electrons and holes are $0.36 m^{2} V^{- 1} s^{- 1}$ and 0.14 $m^{2} V^{- 1} s^{- 1}$ respectively, then the current flowing through the plate will be

1

0.25 A

2

0.45 A

3

0.56 A

4

0.64 A

Explanation:

D Given,
$V = 2 volt$
$l = 0.5 mm = 0.5 \times 10^{- 3} m$
$A = 1 {cm}^{2} = 10^{- 4} m^{2}$
$n = 2 \times 10^{19} / m^{2}$
$μ_{e} = 0.36 m^{2} V^{- 1} s^{- 1}$
$μ_{h} = 0.14 m^{2} V^{- 1} s^{- 1}$
We know conductivity-
$σ = ne (μ_{c} + μ_{h})$
$σ = 2 \times 10^{19} \times 1.6 \times 10^{- 19} (0.36 + 0.14)$
$σ = 1.6 (Ω m)^{- 1}$
Then,
$R = ρ \frac{l}{A} = \frac{l}{σ A} = \frac{0.5 \times 10^{- 3}}{1.6 \times 10^{- 4}} = \frac{25}{8} Ω$ So, $i = \frac{V}{R} = \frac{2}{25 / 8} = 0.64 A$

VITEEE-2010

Semiconductor Electronics Material Devices and Simple Circuits

150559 A conductor and a semi-conductor are connected in parallel as shown in the figure. At a certain voltage both ammeters registers the same current. If the voltage of the DC source is increased then
original image

1 the ammeter connected to the semiconductor will register higher current than the ammeter connected to the conductor

2 the ammeter connected to the conductor will register higher current than the ammeter connected to the semiconductor

3 the ammeters connected to both semiconductor and conductor will register the same current

4 the ammeter connected to both semiconductor and conductor will register no change in the current

Explanation:

C As per given figure, here, conductor and semiconductor are connected in parallel. Therefore, voltage across them will be the same. If ammeters show the same reading then their resistances $R = \frac{V}{I}$ will be the same. If voltage is increased by small value then following the same relation $V \propto I$ for constant $R$ .
Hence, both conductor and semiconductor will display the same currents.

VITEEE-2007

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