831 + Interview Questions in SEMICONDUCTORS IN GENERAL AWARENESS Page 16 InterviewSolution

751.	(a) Name an instrument that measures electric current in a circuit. Define the unit of electric current. (b) What do the following symbols, (Fig. 3.58) mean in a circuit diagram ? (c) An electric circuit consisting of a `0.5 m` long nichrome wire `XY`, an ammeter, a voltmeter, four cells of `1.5 V` each and a plug key was set up. (i) Draw a diagram of this electric circuit to study the relation between the potential difference maintained between the points `X` and `Y` and the electric current flowing through `XY`. (ii) Graph shown in (Fig. 3.59) was plotted `V and I` values. What would be the values of `V//I` ratios when the potential differences are `0.8 V, 1.2 V and 1.6 V` respectively ? What conclusion do you draw from these values ? .A. B. C. D.
Answer» Correct Answer - B

Discussion

752.	In transistor symbols, the arrows shows shows the direction of-A. current in the emitterB. electron current in the emitterC. holes current in the emitterD. None of these
Answer» Correct Answer - A

Discussion

753.	Which one of the following circuits shows correct biasing of a PNP transistor to operate in active region in the CE mode-A. B. C. D.
Answer» Correct Answer - D

Discussion

754.	In transistor symbols, the arrows shows the direction of-A. holes flow in the emitter regionB. electrons flow in the emitter regionC. minority carrier flow in the emitter regionD. majority carrier flow in the emitter region
Answer» Correct Answer - A

Discussion

755.	In an n-p-n transistor, the emitter current is always less than the emitter current, becauseA. equal to the base currentB. slightly more than the collector currentC. equal to the collector currentD. slightly less than the collector current
Answer» Correct Answer - B

Discussion

756.	The output current versus time curve of a rectifire is shown in the figure. The voltage value of output current in this case is
Answer» Correct Answer - C

Discussion

757.	In case of an n-p-n transistor, the collector current is always less than the emitter current, becauseA. emitter side is forward biasedB. collector side is reverse biasedC. a few electrons are lost in the baseD. a few electrons are lost in the emitter
Answer» Correct Answer - C

Discussion

758.	An `NPN`-transistor circuit is arranged as shown in figure. It is A. a common-base amplifier circuitB. a common-emitter amplifier circuitC. a common-collector amplifier circuitD. none of the above.
Answer» Correct Answer - B

Discussion

759.	`npn` transistors are preferred to `pnp` transistors because they haveA. low costB. low dissipation of energyC. capable of handling large powerD. electrons have high mobility than holes and hence high mobility of energy
Answer» Correct Answer - D

Discussion

760.	`npn` transistors are preferred to `pnp` transistors because they haveA. they have low costB. they have low dissipation energyC. they are capable of handling large powerD. electrons have high mobility thn holes.
Answer» Correct Answer - D

Discussion

761.	An audio signal consists of two distinct sound. One a human speech signal in the frequency band of `200Hz` to `2700Hz`, while the other is a high frequency music signal in the frequency band of `10200Hz` to `15200Hz`. The ratio of the AM signal together to the AM signal band width required to send just the human speech is:A. 2B. 3C. 5D. 6
Answer» Correct Answer - D

Discussion

762.	Choose the correct statement.A. In frequency modulation the amplitude of the high frequency carrier wace is made to vary in proportion to the frequency of the audio signalB. In amplitude modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signalC. In amplitude modulation the frequency of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signalD. In frequency modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.
Answer» Correct Answer - B

Discussion

763.	A single of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz. The frequencies of the resultant signal is/are :-A. 2005 kHz,2000 kHz and 1995 kHzB. 2000 kHz and 1995 kHzC. 2 MHz onlyD. 2005 kHz and 1995 kHz
Answer» Correct Answer - A

Discussion

764.	In an unbiased p-n junction electrons diffuse from n-region to p-region because :-A. Holes in p-region attract themB. Electrons travel across the junction due to potential differenceC. Only electrons move from n to p region and not the vice - versaD. Electron concentration n- region is more as compared to that in p-region
Answer» Correct Answer - D

Discussion

765.	In an unbiased p-n junction, holes diffuse from the p-region to n-region because (A) free electrons in the n-region attract them.(B) they move across the junction by the potential difference. (C) hole concentration in p-region is more as compared to n-region. (D) all the above.
Answer» (C) hole concentration in p-region is more as compared to n-region.

Discussion

766.	For the given circuit of `PN`-junction diode, which of the following statements is correct? A. In reverse biasing the voltage across R is 2 VB. In forward biasing the voltage across R is 2 VC. In forward biasing the voltage across R is VD. In reverse biasing the voltage across R is V
Answer» Correct Answer - C For an ideal junction diode, the forward resistance is zero and hence there is no voltage drop across it, when it is forward biased. Hence voltage across R is V. Option (c). But in reverse biasing it does not conduct. It has infinite resistance. `therefore` Voltage across the diode is V and voltage across R is zero.

Discussion

767.	Two ideal diodes are connected to a battery as shown in the circuit. The current supplied by the battery is A. 0.75 AB. zeroC. 0.25 AD. 0.5 A
Answer» Correct Answer - D Here `D_(1)` is in forward bias and `D_(2)` is in reverse bias so `I=(V)/(R)=(5)/(10)=(1)/(2)` AmP.

Discussion

768.	Majority carries in semiconductor areA. holes in n-type and p-type bothB. electron in n-type and p-type bothC. holes in n-type and electron in p-typeD. holes in p-type and electrons in n-type
Answer» Correct Answer - D

Discussion

769.	A pure semiconductorA. has low resistanceB. allows adequate current to pass through itC. is an intrinsic semiconductorD. all of these
Answer» Correct Answer - C

Discussion

770.	If the number of electrons (majority carrier) in a semiconductor is 5 X 10^20 m^-3 and μe is 0.135 mho, find the resistivity of the semiconductor.(a) 0.0926 Ωm(b) 0.0945 Ωm(c) 0.0912 Ωm(d) 0.0978 Ωm
Answer» Correct option is (a) 0.0926 Ωm Explanation: We know, Conductivity, σ = ene μe = 5 X 1.6 X 0.135 X 10 mho/m = 10.8 mho/m Resistivity = 1/σ = 0.0926 Ωm.

Discussion

771.	A pure semiconductor hasA. an infinite resistance at `0^(@)` CB. a finite resistance which does not depend upon temperatureC. a finite resistance which decrease with temperatureD. a finite reststance which increases with temperature
Answer» Correct Answer - C

Discussion

772.	The energy gap between conductionband and valence band is of the order of 0.07 eV. It is a/anA. a conductorB. semiconductorC. an insulatorD. a super conductor
Answer» Correct Answer - B

Discussion

773.	When a battery is connected to a `P`-type semiconductor with a metallic wire, the current in the semiconductor (predominantly), inside the metallic wire and that inside the bettery respectively due toA. Holes, ions, electronsB. Ions, electrons, holesC. Electrons, ions, holesD. Holes, electrons, ions
Answer» Correct Answer - D In the p-type semiconductor, the holes are the majority charge carriers. Inside the conducting wire, free electrons are the charge carriers while in a battery ( cell), ions are the charge carriers. Thus the correct sequence is holes, electrons, ions.

Discussion

774.	The resistivity of a semiconductor at room temperature is in betweenA. `10^(10) ` to `10^(12) Omega ` cmB. `10^(6) ` to `10^(8) Omega` cmC. `10^(-3)` to `10^(6) Omega` cmD. `10^(-2)` to `10^(-5) Omega` cm
Answer» Correct Answer - C

Discussion

775.	When a metal is heated, it emits predominantlyA. mesonsB. protonsC. neutronsD. electrons
Answer» Correct Answer - D

Discussion

776.	The number of electrons in the valence shell of a semiconductor isA. 1B. 2C. 3D. 4
Answer» Correct Answer - D

Discussion

777.	Generally, the number of electrons in the valence shell of good conductors isA. equal to 4B. less than 3C. more than 3D. more than 8
Answer» Correct Answer - D

Discussion

778.	There is no hole current in good conductors, becauseA. they do not have valence bandB. they do not have conduction bandC. their valence and conduction bands overlapD. their valence bands overlap only
Answer» Correct Answer - C

Discussion

779.	There is no hole current in conductors because they haveA. have large forbidden energy gapB. have overlapping valence and conduction bandsC. are full of electron gasD. have no valence band
Answer» Correct Answer - B In good conductors, which are metals there is no gap between valence band and conduction band. Hence, no holes exist.

Discussion

780.	A transistor is used in common emitter configuration. Given its `alpha=0.9`, calculate the change in collector current when the base current changes by `2muA`.A. `3 muA`B. `9 muA`C. `18 muA`D. `30 muA`
Answer» Correct Answer - B `beta=alpha/(1-alpha) = 0.9/(1-0.9) = 9 " " beta=(DeltaI_C)/(DeltaI_B)` `therefore DeltaI_C=beta. DeltaI_B " " therefore DeltaI_C=9xx2=18 muA`

Discussion

781.	A zener diode is to be used as a voltage regulator. Identify the correct set up.A. B. C. D.
Answer» Correct Answer - A

Discussion

782.	The V-I characterisctic for a p-n junction diode is plotted as shown in the figure. From the plot we can conclude that `[V_(b)to` breakdown voltage, `V_(k)to` knee voltage]A. the forward bias resistance of diode is very high, almost infinity for small values of V and after a certain value it becomes very lowB. the reverse bias resistance of diode is very high in the beginning upto breakdown voltage is not achievedC. both forward and reverse bias resistances are same for all voltagesD. both (A) and (B) are correct
Answer» Correct Answer - D

Discussion

783.	The temperature (T) dependence of resistivity (rho) of a semiconductor is represented by :A. Figure (2)B. Figure (3)C. Figure (4)D. Figure (1)
Answer» Correct Answer - B When the temperature is increased, the conductivity of a semiconductor is increased and its resistivity is decreased exponentially. This is shown in figure (3) or option (b).

Discussion

784.	The barrier potential of a `p-n`-junction depends on (i) Type of semiconductor material (ii) Amount of doping (iii) Temperature which of the following is correct?A. (ii) and (iii) onlyB. (i),(ii) and (iii)C. (i) and (ii) onlyD. (ii) only
Answer» Correct Answer - B The barrier potential of a p-n junction depends upon (i), (ii) and (iii).

Discussion

785.	The electrical conductivity of a semiconductor increases when electromagnatic radiation of wavelength shorter than 2480 nm is incident on it. The band gap (in eV) for the semiconductor is `[hc=1242 eV nm]`A. 1.1eVB. 2.5eVC. 0.5 eVD. 0.7 eV
Answer» Correct Answer - C Energy `=12400/(lambda("in" Å)) =12400/24800=0.5 eV`

Discussion

786.	The electrical conductivity of a semiconductor increases when electromagnatic radiation of wavelength shorter than 2480 nm is incident on it. The band gap (in eV) for the semiconductor is `[hc=1242 eV nm]`A. `1.1 eV`B. `2.5 eV`C. `0.5 eV`D. `0.7 eV`
Answer» Correct Answer - C Energy band `= (12400(eV-Å))/(lamda(Å))=(12400)/(24800)=0.5eV`.

Discussion

787.	A hole diffuses from the p-side to the n-side in a p-n junction. this means that.(a) a bond is broken on the n-side and the electron freed from the bond jumps to the conduction band(b) a conduction electron on the p-side jumps to a broken bond to complete it(c) a bond is broken on the n-side and the electron freed from the bond jumps to a broken bond on the p-side to complete it(d) a bond is broken on the p-side and the electron freed from the bond jumps to a broken bond on the n-side to complete it.
Answer» (c) a bond is broken on the n-side and the electron freed from the bond jumps to a broken bond on the p-side to complete it

Discussion

788.	In a p-n junction, the depletion region is 400 nm wide and an electric field of 5 x 105 V/m exists in it. (a) Find the height of the potential barrier, (b) What should be the minimum kinetic energy of a conduction electron which can diffuse from the n-side to the p-side ?
Answer» Depletion region ‘d’ = 400 nm = 4 x10^–7 m Electric field E = 5 x10⁵ V/m a) Potential barrier V = E x d = 0.2 V b) Kinetic energy required = Potential barrier x e = 0.2 eV [Where e = Charge of electron]

788.

In a p-n junction, the depletion region is 400 nm wide and an electric field of 5 x 105 V/m exists in it. (a) Find the height of the potential barrier, (b) What should be the minimum kinetic energy of a conduction electron which can diffuse from the n-side to the p-side ?

Answer»

Depletion region ‘d’ = 400 nm = 4 x10^–7 m

Electric field E = 5 x10⁵ V/m

a) Potential barrier V = E x d = 0.2 V

b) Kinetic energy required = Potential barrier x e = 0.2 eV [Where e = Charge of electron]

Discussion

789.	In a p-n junction, a potential barrier of 250 meV exists across the junction. A hole with a kinetic energy of 300 meV approaches the junction. Find the kinetic energy of the hole when it crosses the junction if the hole approached the junction (a) from the p-side and (b) from the n-side.
Answer» Potential barrier ‘d’ = 250 meV Initial KE of hole = 300 meV We know: KE of the hole decreases when the junction is forward biased and increases when reverse blased in the given ‘Pn’ diode. So, a) Final KE = (300 – 250) meV = 50 meV b) Initial KE = (300 + 250) meV = 550 meV

789.

In a p-n junction, a potential barrier of 250 meV exists across the junction. A hole with a kinetic energy of 300 meV approaches the junction. Find the kinetic energy of the hole when it crosses the junction if the hole approached the junction (a) from the p-side and (b) from the n-side.

Answer»

Potential barrier ‘d’ = 250 meV

Initial KE of hole = 300 meV

We know: KE of the hole decreases when the junction is forward biased and increases when reverse blased in the given ‘Pn’ diode.

So,

a) Final KE = (300 – 250) meV = 50 meV

b) Initial KE = (300 + 250) meV = 550 meV

Discussion

790.	Suppose the energy liberated in the recombination of a hole-electron pair is converted into electromagnetic radiation. If the maximum wavelength emitted is 820 nm, what is the band gap ?
Answer» λ = 820 nm E = hc /λ = 1242/820 = 1.5 eV

790.

Suppose the energy liberated in the recombination of a hole-electron pair is converted into electromagnetic radiation. If the maximum wavelength emitted is 820 nm, what is the band gap ?

Answer»

λ = 820 nm

E = hc /λ = 1242/820 = 1.5 eV

Discussion

791.	The maximum wavelength of electromagnetic radiation, which can create a hole-electron pair in germanium. (Given that forbidden energy gap in germanium is 0.72 eV)A. 172220 ÅB. 172.2 ÅC. 17222 ÅD. 1722 Å
Answer» Correct Answer - C Energy gap, `E_(g)=(hc)/(lambda) and lambda=(hc)/(E_(g))=(12400)/(0.72)=17222`Å

Discussion

792.	Find the maximum wavelength of electromagnetic radiation which can create a hole—electron pair in germanium. The band gap in germanium is 0.65 eV.
Answer» Band Gap = 0.65 eV, λ =? E = hc /λ = 1242 / 0.65 = 1910.7 x 10^–9 m = 1.9 x10^–5 m.

792.

Find the maximum wavelength of electromagnetic radiation which can create a hole—electron pair in germanium. The band gap in germanium is 0.65 eV.

Answer»

Band Gap = 0.65 eV, λ =?

E = hc /λ = 1242 / 0.65 = 1910.7 x 10^–9 m = 1.9 x10^–5 m.

Discussion

793.	The electrical conductivity of pure germanium can be increased byA. increassing the temperatureB. doping acceptor impuritiesC. doping donor impuritiesD. irradiating ultraviolet light on it.
Answer» Correct Answer - A::B::C::D

Discussion

794.	The electrical conductivity of pure germanium can be increased by(a) increased the temperature(b) doping acceptor impurities(c) doping donor impurities(d) irradiating ultraviolet light on it.
Answer» The correct answer is all

Discussion

795.	On which factors does the electrical conductivity of a pure semiconductor depend at a given temperature?
Answer» For pure semiconductor, the number density of free electrons and number density of holes is equal. Thus, at a given temperature, the conductivity of pure semiconductor depends on the number density of charge carriers in the semiconductor.

Discussion

796.	`C` and `Si` both have same lattice structure, having `4` bonding electrons in each. However, `C` is insulator whereas `Si` is intrinsic semiconductor. This is becauseA. in case of C the valence band is not completely filled at absolute zero temperature.B. in case of C the conduction band is partly filled even at absolute zero temperature.C. The four bonding electrons in the case of C ile in the second orbit, whereas in the case of Si they lie in the third.D. The four bonding electrons in the case of C lie in the third orbit, whereas for Si they lie in the fourth orbit.
Answer» Correct Answer - C `.^(6)C=1S^(2),2S^(2)2P^(2)` `.^(14)Si=1S^(2),2S^(2)2P^(6),3S^(2)3P^(2)` As they are away from nucleus, so effect of nucleus is less

Discussion

797.	The electrical conductivity of pure silicon can be increased byA. doping with acceptor impuritiesB. doping with donor impuritiesC. increasing its temperatureD. all the above
Answer» Correct Answer - D

Discussion

798.	A conductor, an insulator, a semiconductor and an alloy are heated by `20^@C` above the room temperature. Then there is an increase in the conductivity of theA. ConductorB. SemiconductorC. InsulatorD. Alloy
Answer» Correct Answer - B

Discussion

799.	In an insulatorA. The valence band is partially filled with electronsB. conduction band is partially filled withelectronsC. conduction band is empty and the valence band is filled with electronsD. conduction band is filled with electrons and valence band empty
Answer» Correct Answer - C

Discussion

800.	In an insulatorA. the valence band is partially filled with electronsB. conduction band is partially filled with electronsC. conduction band is empty and the valence band is filled with electronsD. conduction band is filled with electrons and valence band empty
Answer» Correct Answer - C

Discussion

Explore topic-wise InterviewSolutions in .

In transistor symbols, the arrows shows shows the direction of-A. current in the emitterB. electron current in the emitterC. holes current in the emitterD. None of these

Which one of the following circuits shows correct biasing of a PNP transistor to operate in active region in the CE mode-A. B. C. D.

In transistor symbols, the arrows shows the direction of-A. holes flow in the emitter regionB. electrons flow in the emitter regionC. minority carrier flow in the emitter regionD. majority carrier flow in the emitter region

In an n-p-n transistor, the emitter current is always less than the emitter current, becauseA. equal to the base currentB. slightly more than the collector currentC. equal to the collector currentD. slightly less than the collector current

The output current versus time curve of a rectifire is shown in the figure. The voltage value of output current in this case is

In case of an n-p-n transistor, the collector current is always less than the emitter current, becauseA. emitter side is forward biasedB. collector side is reverse biasedC. a few electrons are lost in the baseD. a few electrons are lost in the emitter

An `NPN`-transistor circuit is arranged as shown in figure. It is A. a common-base amplifier circuitB. a common-emitter amplifier circuitC. a common-collector amplifier circuitD. none of the above.

`npn` transistors are preferred to `pnp` transistors because they haveA. low costB. low dissipation of energyC. capable of handling large powerD. electrons have high mobility than holes and hence high mobility of energy

`npn` transistors are preferred to `pnp` transistors because they haveA. they have low costB. they have low dissipation energyC. they are capable of handling large powerD. electrons have high mobility thn holes.

A single of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz. The frequencies of the resultant signal is/are :-A. 2005 kHz,2000 kHz and 1995 kHzB. 2000 kHz and 1995 kHzC. 2 MHz onlyD. 2005 kHz and 1995 kHz

In an unbiased p-n junction, holes diffuse from the p-region to n-region because (A) free electrons in the n-region attract them.(B) they move across the junction by the potential difference. (C) hole concentration in p-region is more as compared to n-region. (D) all the above.

For the given circuit of `PN`-junction diode, which of the following statements is correct? A. In reverse biasing the voltage across R is 2 VB. In forward biasing the voltage across R is 2 VC. In forward biasing the voltage across R is VD. In reverse biasing the voltage across R is V

Two ideal diodes are connected to a battery as shown in the circuit. The current supplied by the battery is A. 0.75 AB. zeroC. 0.25 AD. 0.5 A

Majority carries in semiconductor areA. holes in n-type and p-type bothB. electron in n-type and p-type bothC. holes in n-type and electron in p-typeD. holes in p-type and electrons in n-type

A pure semiconductorA. has low resistanceB. allows adequate current to pass through itC. is an intrinsic semiconductorD. all of these

If the number of electrons (majority carrier) in a semiconductor is 5 X 10^20 m^-3 and μe is 0.135 mho, find the resistivity of the semiconductor.(a) 0.0926 Ωm(b) 0.0945 Ωm(c) 0.0912 Ωm(d) 0.0978 Ωm

A pure semiconductor hasA. an infinite resistance at `0^(@)` CB. a finite resistance which does not depend upon temperatureC. a finite resistance which decrease with temperatureD. a finite reststance which increases with temperature

The energy gap between conductionband and valence band is of the order of 0.07 eV. It is a/anA. a conductorB. semiconductorC. an insulatorD. a super conductor

The resistivity of a semiconductor at room temperature is in betweenA. `10^(10) ` to `10^(12) Omega ` cmB. `10^(6) ` to `10^(8) Omega` cmC. `10^(-3)` to `10^(6) Omega` cmD. `10^(-2)` to `10^(-5) Omega` cm

When a metal is heated, it emits predominantlyA. mesonsB. protonsC. neutronsD. electrons

The number of electrons in the valence shell of a semiconductor isA. 1B. 2C. 3D. 4

Generally, the number of electrons in the valence shell of good conductors isA. equal to 4B. less than 3C. more than 3D. more than 8

There is no hole current in good conductors, becauseA. they do not have valence bandB. they do not have conduction bandC. their valence and conduction bands overlapD. their valence bands overlap only

There is no hole current in conductors because they haveA. have large forbidden energy gapB. have overlapping valence and conduction bandsC. are full of electron gasD. have no valence band

A transistor is used in common emitter configuration. Given its `alpha=0.9`, calculate the change in collector current when the base current changes by `2muA`.A. `3 muA`B. `9 muA`C. `18 muA`D. `30 muA`

A zener diode is to be used as a voltage regulator. Identify the correct set up.A. B. C. D.

The temperature (T) dependence of resistivity (rho) of a semiconductor is represented by :A. Figure (2)B. Figure (3)C. Figure (4)D. Figure (1)

The barrier potential of a `p-n`-junction depends on (i) Type of semiconductor material (ii) Amount of doping (iii) Temperature which of the following is correct?A. (ii) and (iii) onlyB. (i),(ii) and (iii)C. (i) and (ii) onlyD. (ii) only

The electrical conductivity of a semiconductor increases when electromagnatic radiation of wavelength shorter than 2480 nm is incident on it. The band gap (in eV) for the semiconductor is `[hc=1242 eV nm]`A. 1.1eVB. 2.5eVC. 0.5 eVD. 0.7 eV

The electrical conductivity of a semiconductor increases when electromagnatic radiation of wavelength shorter than 2480 nm is incident on it. The band gap (in eV) for the semiconductor is `[hc=1242 eV nm]`A. `1.1 eV`B. `2.5 eV`C. `0.5 eV`D. `0.7 eV`

In a p-n junction, the depletion region is 400 nm wide and an electric field of 5 x 105 V/m exists in it. (a) Find the height of the potential barrier, (b) What should be the minimum kinetic energy of a conduction electron which can diffuse from the n-side to the p-side ?

In a p-n junction, a potential barrier of 250 meV exists across the junction. A hole with a kinetic energy of 300 meV approaches the junction. Find the kinetic energy of the hole when it crosses the junction if the hole approached the junction (a) from the p-side and (b) from the n-side.

Suppose the energy liberated in the recombination of a hole-electron pair is converted into electromagnetic radiation. If the maximum wavelength emitted is 820 nm, what is the band gap ?

The maximum wavelength of electromagnetic radiation, which can create a hole-electron pair in germanium. (Given that forbidden energy gap in germanium is 0.72 eV)A. 172220 ÅB. 172.2 ÅC. 17222 ÅD. 1722 Å

Find the maximum wavelength of electromagnetic radiation which can create a hole—electron pair in germanium. The band gap in germanium is 0.65 eV.

The electrical conductivity of pure germanium can be increased byA. increassing the temperatureB. doping acceptor impuritiesC. doping donor impuritiesD. irradiating ultraviolet light on it.

The electrical conductivity of pure germanium can be increased by(a) increased the temperature(b) doping acceptor impurities(c) doping donor impurities(d) irradiating ultraviolet light on it.

On which factors does the electrical conductivity of a pure semiconductor depend at a given temperature?

The electrical conductivity of pure silicon can be increased byA. doping with acceptor impuritiesB. doping with donor impuritiesC. increasing its temperatureD. all the above

A conductor, an insulator, a semiconductor and an alloy are heated by `20^@C` above the room temperature. Then there is an increase in the conductivity of theA. ConductorB. SemiconductorC. InsulatorD. Alloy

In an insulatorA. The valence band is partially filled with electronsB. conduction band is partially filled withelectronsC. conduction band is empty and the valence band is filled with electronsD. conduction band is filled with electrons and valence band empty

In an insulatorA. the valence band is partially filled with electronsB. conduction band is partially filled with electronsC. conduction band is empty and the valence band is filled with electronsD. conduction band is filled with electrons and valence band empty