Practice Start Test

Semiconductor

NEET Physics · 93 questions · Page 7 of 10 · Click an option or "Show Solution" to reveal answer

Q61
A transistor is operated in common-emitter configuration at V C = 2V such that a change in the base current from 100 μ\mu A to 200 μ\mu A produces a change in the collector current from 5 mA to 10 mA. The current gain is
A 100
B 150
C 50
D 75
Correct Answer
Option C
Solution

Current gain, β\beta =

ΔICΔIB{{\Delta {I_C}} \over {\Delta {I_B}}}

=

(10−5)×10−3(200−100)×10−6{{\left( {10 - 5} \right) \times {{10}^{ - 3}}} \over {\left( {200 - 100} \right) \times {{10}^{ - 6}}}}

= 50

Q62
A p-n photodiode is fabricated from a semiconductor with a band gap of 2.5 eV. It can detect a signal of wavelength
A 4000 nm
B 6000 nm
C 4000 A∘\mathop A\limits^ \circ
D 6000 A∘\mathop A\limits^ \circ
Correct Answer
Option C
Solution

λ\lambda max =

hcE{{hc} \over E}

=

6.6×10−34×3×1082.5×1.6×10−19{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {2.5 \times 1.6 \times {{10}^{ - 19}}}}

= 5000

Ao\mathop A\limits^o

Now the wavelength detected by photodiode be less than λ\lambda max , hence it can detect a signal of wave length 4000

Ao\mathop A\limits^o

.

Q63
Sodium has body centred packing. Distance between two nearest atoms is 3.7 A∘\mathop A\limits^ \circ . The lattice parameter is
A 4.3 A∘\mathop A\limits^ \circ
B 3.0 A∘\mathop A\limits^ \circ
C 8.6 A∘\mathop A\limits^ \circ
D 6.8 A∘\mathop A\limits^ \circ
Correct Answer
Option A
Solution

Distance between nearest atoms in body centred cubic lattice (bcc), d =

32a{{\sqrt 3 } \over 2}a

Given d = 3.7

A∘\mathop A\limits^ \circ

∴\therefore

a=3.7×23a = {{3.7 \times 2} \over {\sqrt 3 }}

= 4.3

A∘\mathop A\limits^ \circ
Q64
The voltage gain of an amplifier with 9% negative feedback is 10. The voltage gain without feedback will be
A 1.25
B 100
C 90
D 10
Correct Answer
Option B
Solution

Negative feedback is applied to reduce the output voltage of an amplifier.

If there is no negative feedback, the value of output voltage could be very high.

In the options given, the maximum value of voltage gain is 100.

Hence it is the correct option.

Q65
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
A 1 × \times 10 14 Hz
B 20 × \times 10 14 Hz
C 10 × \times 10 14 Hz
D 5 × \times 10 14 Hz
Correct Answer
Option D
Solution

Energy gap E g = 2.0 eV E g = 2.0 × 1.6 × 10 –19 J E g = hf f = frequency of radiation ∴\therefore f =

Egh{{{E_g}} \over h}

=

3.2×10−196.625×10−34{{3.2 \times {{10}^{ - 19}}} \over {6.625 \times {{10}^{ - 34}}}}

= 5 ×\times 10 14 Hz

Q66
If the lattice parameter for a crystalline structure is 3.6 A∘\mathop A\limits^ \circ , then the atomic radius in fcc crystal is
A 2.92 A∘\mathop A\limits^ \circ
B 1.27 A∘\mathop A\limits^ \circ
C 1.81 A∘\mathop A\limits^ \circ
D 2.10 A∘\mathop A\limits^ \circ
Correct Answer
Option B
Solution

The atomic radius in a f.c.c. crystal is =

a22{a \over {2\sqrt 2 }}

∴\therefore Atomic radius =

3.622{{3.6} \over {2\sqrt 2 }}

= 1.27

A∘\mathop A\limits^ \circ
Q67
For a cubic crystal structure which one of the following relations indicating the cell characteristics is correct ?
A a ≠ \ne b ≠ \ne c and α\alpha = β\beta = γ\gamma = 90 o
B a = b = c and α\alpha ≠ \ne β\beta ≠ \ne γ\gamma = 90 o
C a = b = c and α\alpha = β\beta = γ\gamma = 90 o
D a ≠ \ne b ≠ \ne c and ;α\alpha ≠ \ne β\beta ≠ \ne and γ\gamma = 90 o
Correct Answer
Option C
Solution

In a cubic crystal structure a = b = c and α\alpha = β\beta = γ\gamma = 90 o

Q68
In the energy band diagram of a material shown below, the open circles and filled circles denote holes and electrons respectively. The material is
A an insulator
B a metal
C an n-type semiconductor
D a p-type semiconductor.
Correct Answer
Option D
Solution

For a p-type semiconductor, the acceptor energy level, as shown in the diagram, is slightly above the top E v of the valence band.

With very small supply of energy an electron from the valence band can jump to the level E C and ionise acceptor negatively.

Q69
In the following circuit, the output Y for all possible inputs A and B is expressed by the truth table.
A ABY001011101110\begin{array}{lll}A & B & Y \\ 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}
B ABY001010100110\begin{array}{lll}A & B & Y \\ 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ 1 & 1 & 0 \end{array}
C ABY000011101111\begin{array}{lll}A & B & Y \\ 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 1 \end{array}
D ABY000010100111\begin{array}{lll}A & B & Y \\ 0 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \end{array}
Correct Answer
Option C
Solution

Y' =

A+B‾\overline {A + B}

Y =

Y′‾\overline {Y'}

Y =

A+B‾‾\overline{\overline {A + B}}

Y = A + B which is OR-gate.

Q70
The following figure shows a logic gate circuit with two inputs A and B and the output C. The voltage waveforms of A, B and C are as shown below. The logic circuit gate is
A OR gate
B AND gate
C NAND gate
D NOR gate
Correct Answer
Option B
Solution

The truth table corresponding to waveform is given by

ABC111010100000\begin{array}{lll}A & B & C \\ 1 & 1 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{array}

∴\therefore The given logic circuit gate is AND gate.

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