QBManager

Semiconductor Electronics Material Devices and Simple Circuits

150628 What is the maximum wavelength of photon that would excite an electron in the valence band of diamond to the conduction band? The energy gap for diamond is $5.5 eV$ .

1

169 nm

2

205 nm

3

226 nm

4

350 nm

Explanation:

C Given
Energy gap of diamond $(E) = 5.5 eV = 5.5 \times 1.6 \times 10^{- 19} J$
We know that,
$E = \frac{h c}{λ}$
$λ = \frac{h c}{E}$
(Planks constant $h = 6.63 \times 10^{- 34}, C = 3 \times 10^{8} m / s$ )
$λ = \frac{6.63 \times 10^{- 34} \times 3 \times 10^{8}}{5.5 \times 1.6 \times 10^{- 19}}$
$λ = 2.26 \times 10^{- 7}$
$λ = 226 nm$

AMU-2014

Semiconductor Electronics Material Devices and Simple Circuits

150632 You are given four semiconductors $P, Q, R$ and $S$ with respective band gaps $4 eV, 3 eV, 2 eV$ and $1 eV$ for use in a photodetector to detect $λ = 1400 nm$ . Select the suitable semiconductor.

1

P

2

Q

3

R

4

S

Explanation:

D Given,
Energy band gap $(E_{g})$ of semiconductors $P, Q, R$ and $S$ are $4 eV, 3 eV, 2 eV$ and $1 eV$ respectively.
Wavelength $(λ) = 1400 nm = 1400 \times 10^{- 9} m$
Energy of photon $E = \frac{h c}{λ}$
$E = \frac{6.63 \times 10^{- 34} \times 3 \times 10^{8}}{1400 \times 10^{- 9} \times 1.6 \times 10^{- 19}}$
$E = 0.89 eV \approx 1 eV$ So, suitable semiconductor is $S$ with band gap $1 eV$ .

AMU-2013

Semiconductor Electronics Material Devices and Simple Circuits

150638 An intrinsic semiconductor has a resistivity of $0.50 Ω$ at room temperature. Find the intrinsic carrier concentration if the mobilities of electrons and holes are $0.39 m^{2} /$ volt sec and $0.11 m^{2} /$ volt sec respectively

1

1.2 \times 10^{18} / m^{3}

2

2.5 \times 10^{19} / m^{3}

3

1.9 \times 10^{20} / m^{3}

4

3.1 \times 10^{21} / m^{3}

Explanation:

B Given,
Resistivity $(ρ) = 0.50 Ω$
Then conductivity,
$σ = \frac{1}{ρ} = \frac{1}{0.50}$
$σ = 2$
$μ_{c} = 0.39 m^{2} / volt$
$μ_{p} = 0.11 m^{2} / volt$
Then we know that,
$σ = e (n μ_{e} + n μ_{p})$
$σ = en (μ_{e} + μ_{p})$
$2 = en (0.39 + 0.11)$
$n = \frac{2}{0.5 \times 1.6 \times 10^{- 19}}$
$n = 2.5 \times 10^{19} / m^{3}$

AMU-2011

Semiconductor Electronics Material Devices and Simple Circuits

150662 Suppose a pure $Si$ crystal has $5 \times 10^{28}$ atoms $m^{- 3}$ . It is doped by 1 ppm concentration of pentavalent As. Calculate the number of electron and holes. Given that $n_{i} = 1.5 \times 10^{16}$ $m^{- 3}$

1

6.5 \times 10^{9} m^{- 3}

2

4.5 \times 10^{9} m^{- 3}

3

5.5 \times 10^{9} m^{- 3}

4

5.5 \times 10^{9} m^{- 3}

Explanation:

B Given that,
1atom have silicon doped out of atom $10^{6}$
$n_{i} = 1.5 \times 10^{16} m^{- 3}$
Therefore in $5 \times 10^{28}$ atom net doped $= \frac{5 \times 10^{28}}{10^{6}}$
$= 5 \times 10^{22} = n_{h}$
Therefore, $n_{e} \times n_{h} = n_{i}^{2}$
(Einstein equation)
$n_{e} = \frac{n_{i}^{2}}{n_{h}} = \frac{{(1.5 \times 10^{16})}^{2}}{5 \times 10^{22}}$
$n_{e} = 4.5 \times 10^{9} m^{- 3}$ Then, $n_{e} > n_{h}$ so it is $n$ type semi conductor.

GUJCET 18.04.2022

Semiconductor Electronics Material Devices and Simple Circuits

150628 What is the maximum wavelength of photon that would excite an electron in the valence band of diamond to the conduction band? The energy gap for diamond is $5.5 eV$ .

1

169 nm

2

205 nm

3

226 nm

4

350 nm

Explanation:

C Given
Energy gap of diamond $(E) = 5.5 eV = 5.5 \times 1.6 \times 10^{- 19} J$
We know that,
$E = \frac{h c}{λ}$
$λ = \frac{h c}{E}$
(Planks constant $h = 6.63 \times 10^{- 34}, C = 3 \times 10^{8} m / s$ )
$λ = \frac{6.63 \times 10^{- 34} \times 3 \times 10^{8}}{5.5 \times 1.6 \times 10^{- 19}}$
$λ = 2.26 \times 10^{- 7}$
$λ = 226 nm$

AMU-2014

Semiconductor Electronics Material Devices and Simple Circuits

150632 You are given four semiconductors $P, Q, R$ and $S$ with respective band gaps $4 eV, 3 eV, 2 eV$ and $1 eV$ for use in a photodetector to detect $λ = 1400 nm$ . Select the suitable semiconductor.

1

P

2

Q

3

R

4

S

Explanation:

D Given,
Energy band gap $(E_{g})$ of semiconductors $P, Q, R$ and $S$ are $4 eV, 3 eV, 2 eV$ and $1 eV$ respectively.
Wavelength $(λ) = 1400 nm = 1400 \times 10^{- 9} m$
Energy of photon $E = \frac{h c}{λ}$
$E = \frac{6.63 \times 10^{- 34} \times 3 \times 10^{8}}{1400 \times 10^{- 9} \times 1.6 \times 10^{- 19}}$
$E = 0.89 eV \approx 1 eV$ So, suitable semiconductor is $S$ with band gap $1 eV$ .

AMU-2013

Semiconductor Electronics Material Devices and Simple Circuits

150638 An intrinsic semiconductor has a resistivity of $0.50 Ω$ at room temperature. Find the intrinsic carrier concentration if the mobilities of electrons and holes are $0.39 m^{2} /$ volt sec and $0.11 m^{2} /$ volt sec respectively

1

1.2 \times 10^{18} / m^{3}

2

2.5 \times 10^{19} / m^{3}

3

1.9 \times 10^{20} / m^{3}

4

3.1 \times 10^{21} / m^{3}

Explanation:

B Given,
Resistivity $(ρ) = 0.50 Ω$
Then conductivity,
$σ = \frac{1}{ρ} = \frac{1}{0.50}$
$σ = 2$
$μ_{c} = 0.39 m^{2} / volt$
$μ_{p} = 0.11 m^{2} / volt$
Then we know that,
$σ = e (n μ_{e} + n μ_{p})$
$σ = en (μ_{e} + μ_{p})$
$2 = en (0.39 + 0.11)$
$n = \frac{2}{0.5 \times 1.6 \times 10^{- 19}}$
$n = 2.5 \times 10^{19} / m^{3}$

AMU-2011

Semiconductor Electronics Material Devices and Simple Circuits

150662 Suppose a pure $Si$ crystal has $5 \times 10^{28}$ atoms $m^{- 3}$ . It is doped by 1 ppm concentration of pentavalent As. Calculate the number of electron and holes. Given that $n_{i} = 1.5 \times 10^{16}$ $m^{- 3}$

1

6.5 \times 10^{9} m^{- 3}

2

4.5 \times 10^{9} m^{- 3}

3

5.5 \times 10^{9} m^{- 3}

4

5.5 \times 10^{9} m^{- 3}

Explanation:

B Given that,
1atom have silicon doped out of atom $10^{6}$
$n_{i} = 1.5 \times 10^{16} m^{- 3}$
Therefore in $5 \times 10^{28}$ atom net doped $= \frac{5 \times 10^{28}}{10^{6}}$
$= 5 \times 10^{22} = n_{h}$
Therefore, $n_{e} \times n_{h} = n_{i}^{2}$
(Einstein equation)
$n_{e} = \frac{n_{i}^{2}}{n_{h}} = \frac{{(1.5 \times 10^{16})}^{2}}{5 \times 10^{22}}$
$n_{e} = 4.5 \times 10^{9} m^{- 3}$ Then, $n_{e} > n_{h}$ so it is $n$ type semi conductor.

GUJCET 18.04.2022

Semiconductor Electronics Material Devices and Simple Circuits

150628 What is the maximum wavelength of photon that would excite an electron in the valence band of diamond to the conduction band? The energy gap for diamond is $5.5 eV$ .

1

169 nm

2

205 nm

3

226 nm

4

350 nm

Explanation:

C Given
Energy gap of diamond $(E) = 5.5 eV = 5.5 \times 1.6 \times 10^{- 19} J$
We know that,
$E = \frac{h c}{λ}$
$λ = \frac{h c}{E}$
(Planks constant $h = 6.63 \times 10^{- 34}, C = 3 \times 10^{8} m / s$ )
$λ = \frac{6.63 \times 10^{- 34} \times 3 \times 10^{8}}{5.5 \times 1.6 \times 10^{- 19}}$
$λ = 2.26 \times 10^{- 7}$
$λ = 226 nm$

AMU-2014

Semiconductor Electronics Material Devices and Simple Circuits

150632 You are given four semiconductors $P, Q, R$ and $S$ with respective band gaps $4 eV, 3 eV, 2 eV$ and $1 eV$ for use in a photodetector to detect $λ = 1400 nm$ . Select the suitable semiconductor.

1

P

2

Q

3

R

4

S

Explanation:

D Given,
Energy band gap $(E_{g})$ of semiconductors $P, Q, R$ and $S$ are $4 eV, 3 eV, 2 eV$ and $1 eV$ respectively.
Wavelength $(λ) = 1400 nm = 1400 \times 10^{- 9} m$
Energy of photon $E = \frac{h c}{λ}$
$E = \frac{6.63 \times 10^{- 34} \times 3 \times 10^{8}}{1400 \times 10^{- 9} \times 1.6 \times 10^{- 19}}$
$E = 0.89 eV \approx 1 eV$ So, suitable semiconductor is $S$ with band gap $1 eV$ .

AMU-2013

Semiconductor Electronics Material Devices and Simple Circuits

150638 An intrinsic semiconductor has a resistivity of $0.50 Ω$ at room temperature. Find the intrinsic carrier concentration if the mobilities of electrons and holes are $0.39 m^{2} /$ volt sec and $0.11 m^{2} /$ volt sec respectively

1

1.2 \times 10^{18} / m^{3}

2

2.5 \times 10^{19} / m^{3}

3

1.9 \times 10^{20} / m^{3}

4

3.1 \times 10^{21} / m^{3}

Explanation:

B Given,
Resistivity $(ρ) = 0.50 Ω$
Then conductivity,
$σ = \frac{1}{ρ} = \frac{1}{0.50}$
$σ = 2$
$μ_{c} = 0.39 m^{2} / volt$
$μ_{p} = 0.11 m^{2} / volt$
Then we know that,
$σ = e (n μ_{e} + n μ_{p})$
$σ = en (μ_{e} + μ_{p})$
$2 = en (0.39 + 0.11)$
$n = \frac{2}{0.5 \times 1.6 \times 10^{- 19}}$
$n = 2.5 \times 10^{19} / m^{3}$

AMU-2011

Semiconductor Electronics Material Devices and Simple Circuits

150662 Suppose a pure $Si$ crystal has $5 \times 10^{28}$ atoms $m^{- 3}$ . It is doped by 1 ppm concentration of pentavalent As. Calculate the number of electron and holes. Given that $n_{i} = 1.5 \times 10^{16}$ $m^{- 3}$

1

6.5 \times 10^{9} m^{- 3}

2

4.5 \times 10^{9} m^{- 3}

3

5.5 \times 10^{9} m^{- 3}

4

5.5 \times 10^{9} m^{- 3}

Explanation:

B Given that,
1atom have silicon doped out of atom $10^{6}$
$n_{i} = 1.5 \times 10^{16} m^{- 3}$
Therefore in $5 \times 10^{28}$ atom net doped $= \frac{5 \times 10^{28}}{10^{6}}$
$= 5 \times 10^{22} = n_{h}$
Therefore, $n_{e} \times n_{h} = n_{i}^{2}$
(Einstein equation)
$n_{e} = \frac{n_{i}^{2}}{n_{h}} = \frac{{(1.5 \times 10^{16})}^{2}}{5 \times 10^{22}}$
$n_{e} = 4.5 \times 10^{9} m^{- 3}$ Then, $n_{e} > n_{h}$ so it is $n$ type semi conductor.

GUJCET 18.04.2022

Semiconductor Electronics Material Devices and Simple Circuits

150628 What is the maximum wavelength of photon that would excite an electron in the valence band of diamond to the conduction band? The energy gap for diamond is $5.5 eV$ .

1

169 nm

2

205 nm

3

226 nm

4

350 nm

Explanation:

C Given
Energy gap of diamond $(E) = 5.5 eV = 5.5 \times 1.6 \times 10^{- 19} J$
We know that,
$E = \frac{h c}{λ}$
$λ = \frac{h c}{E}$
(Planks constant $h = 6.63 \times 10^{- 34}, C = 3 \times 10^{8} m / s$ )
$λ = \frac{6.63 \times 10^{- 34} \times 3 \times 10^{8}}{5.5 \times 1.6 \times 10^{- 19}}$
$λ = 2.26 \times 10^{- 7}$
$λ = 226 nm$

AMU-2014

Semiconductor Electronics Material Devices and Simple Circuits

150632 You are given four semiconductors $P, Q, R$ and $S$ with respective band gaps $4 eV, 3 eV, 2 eV$ and $1 eV$ for use in a photodetector to detect $λ = 1400 nm$ . Select the suitable semiconductor.

1

P

2

Q

3

R

4

S

Explanation:

D Given,
Energy band gap $(E_{g})$ of semiconductors $P, Q, R$ and $S$ are $4 eV, 3 eV, 2 eV$ and $1 eV$ respectively.
Wavelength $(λ) = 1400 nm = 1400 \times 10^{- 9} m$
Energy of photon $E = \frac{h c}{λ}$
$E = \frac{6.63 \times 10^{- 34} \times 3 \times 10^{8}}{1400 \times 10^{- 9} \times 1.6 \times 10^{- 19}}$
$E = 0.89 eV \approx 1 eV$ So, suitable semiconductor is $S$ with band gap $1 eV$ .

AMU-2013

Semiconductor Electronics Material Devices and Simple Circuits

150638 An intrinsic semiconductor has a resistivity of $0.50 Ω$ at room temperature. Find the intrinsic carrier concentration if the mobilities of electrons and holes are $0.39 m^{2} /$ volt sec and $0.11 m^{2} /$ volt sec respectively

1

1.2 \times 10^{18} / m^{3}

2

2.5 \times 10^{19} / m^{3}

3

1.9 \times 10^{20} / m^{3}

4

3.1 \times 10^{21} / m^{3}

Explanation:

B Given,
Resistivity $(ρ) = 0.50 Ω$
Then conductivity,
$σ = \frac{1}{ρ} = \frac{1}{0.50}$
$σ = 2$
$μ_{c} = 0.39 m^{2} / volt$
$μ_{p} = 0.11 m^{2} / volt$
Then we know that,
$σ = e (n μ_{e} + n μ_{p})$
$σ = en (μ_{e} + μ_{p})$
$2 = en (0.39 + 0.11)$
$n = \frac{2}{0.5 \times 1.6 \times 10^{- 19}}$
$n = 2.5 \times 10^{19} / m^{3}$

AMU-2011

Semiconductor Electronics Material Devices and Simple Circuits

150662 Suppose a pure $Si$ crystal has $5 \times 10^{28}$ atoms $m^{- 3}$ . It is doped by 1 ppm concentration of pentavalent As. Calculate the number of electron and holes. Given that $n_{i} = 1.5 \times 10^{16}$ $m^{- 3}$

1

6.5 \times 10^{9} m^{- 3}

2

4.5 \times 10^{9} m^{- 3}

3

5.5 \times 10^{9} m^{- 3}

4

5.5 \times 10^{9} m^{- 3}

Explanation:

B Given that,
1atom have silicon doped out of atom $10^{6}$
$n_{i} = 1.5 \times 10^{16} m^{- 3}$
Therefore in $5 \times 10^{28}$ atom net doped $= \frac{5 \times 10^{28}}{10^{6}}$
$= 5 \times 10^{22} = n_{h}$
Therefore, $n_{e} \times n_{h} = n_{i}^{2}$
(Einstein equation)
$n_{e} = \frac{n_{i}^{2}}{n_{h}} = \frac{{(1.5 \times 10^{16})}^{2}}{5 \times 10^{22}}$
$n_{e} = 4.5 \times 10^{9} m^{- 3}$ Then, $n_{e} > n_{h}$ so it is $n$ type semi conductor.

GUJCET 18.04.2022