If the slope of the characteristics curve of a vacuum tube diode corresponding to some point is ${10^{ - 3}}{\rm{mA}}{{\rm{V}}^{ - 1}}$, then the plate resistance of diode and its nature are respectively

$1000{\rm{\;k\Omega }}$, dynamic

$100{\rm{\;k\Omega }}$, dynamic

$1000{\rm{\;k\Omega }}$, static

SEMI CONDUCTOR DEVICES MCQs for NEET UG PHYSICS — Medium, Page 2

31 Medium

Remembering +4 | -1

In

CE

mode, the input characteristics of a transistor is the variation of

View answer and explanation

Correct option: A

{I_B}

against

{V_{BE}}

at constant

{V_{CE}}

Option A (

{I_B}

against

{V_{BE}}

at constant

{V_{CE}}

) is correct because the input characteristics of a transistor in CE mode show the relationship between base current (

I_B

) and base-emitter voltage (

V_{BE}

) at a constant collector-emitter voltage (

V_{CE}

).

Solution approach

Correct Answer + Core Reason: ${I_B}$ against ${V_{BE}}$ at constant ${V_{CE}}$ . This is the definition of input characteristics in CE mode.

Step-by-Step Solution:

Step 1: Understand CE Mode: In Common Emitter (CE) mode, the emitter is common to both the input and output circuits.

Step 2: Define Input Characteristics: Input characteristics show the relationship between input current and input voltage at a constant output voltage.

Step 3: Identify Input Current and Voltage: In CE mode, the input current is the base current ( $I_B$ ), and the input voltage is the base-emitter voltage ( $V_{BE}$ ).

Step 4: Identify Constant Output Voltage: The output voltage is the collector-emitter voltage ( $V_{CE}$ ).

Step 5: Conclusion: The input characteristics are the variation of $I_B$ against $V_{BE}$ at constant $V_{CE}$ .

Why other options are wrong:
- Option B: This describes the output characteristics, not the input characteristics.
- Option C: This is not a standard characteristic curve.
- Option D: This is not a standard characteristic curve.

Key Concepts:
- Transistor characteristics
- Common Emitter (CE) mode

Memory Aids:
- Input - Base, Output - Collector

Quick Tips:
- Remember the input and output parameters for each transistor configuration.

32 Medium

32 of 85Question permalink

Analyzing +4 | -1

Statement I: If forward current changes by

1.5 \mathrm{~mA}

when forward voltage in semiconductor diode is changed from

0.5 \mathrm{~V}

to

2 \mathrm{~V}

, the forward resistance of diode will be

1 \Omega

.

Statement II: The forward resistance is given by

R_f=\frac{\Delta V_f}{\Delta I_f}

View answer and explanation

Correct option: D

Statement I is false; Statement II is true

Option D is correct because Statement I is false (the forward resistance is not 1

{\Omega}

) and Statement II is true (the formula for forward resistance is correct).

Solution approach

Correct Answer + Core Reason: Statement I is false; Statement II is true. The forward resistance calculation is incorrect, but the formula is correct.

Step-by-Step Solution:

Step 1: Identify the given information: $\Delta I_f = 1.5 \text{{ mA}} = 1.5 \times 10^{-3} \text{{ A}}$ , $\Delta V_f = 2 \text{{ V}} - 0.5 \text{{ V}} = 1.5 \text{{ V}}$

Step 2: Use the formula for forward resistance: $R_f = \frac{{\Delta V_f}}{{\Delta I_f}}$

Step 3: Substitute the values: $R_f = \frac{{1.5 \text{{ V}}}}{{1.5 \times 10^{-3} \text{{ A}}}}$

Step 4: Calculate the forward resistance: $R_f = 1000 \Omega$

Step 5: Evaluate Statement I: Statement I is false because the forward resistance is 1000 ${\Omega}$ , not 1 ${\Omega}$ .

Step 6: Evaluate Statement II: Statement II is true because the formula for forward resistance is correct.

Why other options are wrong:
- Option A: Statement I is not true.
- Option B: Statement I is not true.
- Option C: Statement II is true, so this option is incorrect.

Key Concepts:
- Forward resistance of a diode

Memory Aids:
- Remember the formula for forward resistance.

Quick Tips:
- Always convert units to SI units before substituting into formulas.

33 Medium

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Application +4 | -1

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

View answer and explanation

Correct option: B

5 \times {10^{14}}

Hz

Option B (

5 \times {10^{14}}

Hz) is correct because the minimum frequency is calculated using

E = hf

, where E is the band gap energy.

Solution approach

Correct Answer + Core Reason: $5 \times {10^{14}}$ Hz. The minimum frequency is calculated using the energy-frequency relationship.

Step-by-Step Solution:

Step 1: Identify the given information: Band gap energy $E = 2.0 \text{{ eV}}$ , Planck's constant $h = 6.6 \times 10^{-34} \text{{ Js}}$

Step 2: Convert the band gap energy to Joules: $E = 2.0 \text{{ eV}} \times 1.6 \times 10^{-19} \text{{ J/eV}} = 3.2 \times 10^{-19} \text{{ J}}$

Step 3: Use the formula for energy of a photon: $E = hf$

Step 4: Solve for frequency: $f = \frac{E}{h}$

Step 5: Substitute the values: $f = \frac{{3.2 \times 10^{-19} \text{{ J}}}}{{6.6 \times 10^{-34} \text{{ Js}}}}$

Step 6: Calculate the frequency: $f \approx 4.85 \times 10^{14} \text{{ Hz}} \approx 5 \times 10^{14} \text{{ Hz}}$

Why other options are wrong:
- **Option A ( $10 \times {10^{14}}$ Hz): Incorrect calculation or wrong formula used.
- Option C ( $1 \times {10^{14}}$ Hz): Incorrect calculation or wrong formula used.
- Option D ( $20 \times {10^{14}}$ Hz): Incorrect calculation or wrong formula used.

Key Concepts:**
- Energy of a photon: $E = hf$
- Conversion between electron volts and Joules

Memory Aids:
- Remember the formula $E = hf$

Quick Tips:
- Always convert units to SI units before substituting into formulas.

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Analyzing +4 | -1

Assertion: The resistivity of a semi-conductor increases with temperture.

Reason: the atoms of semi conductor vibrate with large amplitude at higher temperture thereby increasing its resistivity.

View answer and explanation

Correct option: D

Assertion and reason both are false

Option D (Assertion and reason both are false) is correct because the resistivity of a semiconductor decreases with temperature, and the reason provided is not the primary cause.

Solution approach

Correct Answer + Core Reason: Assertion and reason both are false. The resistivity of a semiconductor decreases with temperature, and the reason provided is not the primary cause.

Step-by-Step Solution:

Step 1: Analyze Assertion: The resistivity of a semi-conductor increases with temperature. This is false. The resistivity of a semiconductor decreases with temperature because more charge carriers are generated.

Step 2: Analyze Reason: The atoms of semi conductor vibrate with large amplitude at higher temperature thereby increasing its resistivity. This is also false. While lattice vibrations do increase, the dominant effect is the increase in charge carriers, which reduces resistivity.

Step 3: Determine the correctness of the statements: Both Assertion and Reason are false.

Final Answer: Assertion and reason both are false

Why other options are wrong:
- Option A: Both Assertion and Reason are false.
- Option B: Both Assertion and Reason are false.
- Option C: Assertion is false.

Key Concepts:
- Resistivity of a semiconductor
- Temperature dependence
- Charge carriers

Memory Aids:
- Semiconductors: Resistivity decreases with increasing temperature.

Quick Tips:
- Remember that increasing temperature in a semiconductor increases the number of charge carriers, thus decreasing resistivity.

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Understanding +4 | -1

A piece of copper and another of germanium are cooled from room temperature (300 K) to 280 K, the resistance of

View answer and explanation

Correct option: A

Copper decreases and germanium increases

Option A (Copper decreases and germanium increases) is correct because the resistance of copper (a metal) decreases with decreasing temperature, while the resistance of germanium (a semiconductor) increases.

Solution approach

Correct Answer + Core Reason: Copper decreases and germanium increases. Metals decrease in resistance, semiconductors increase.

Step-by-Step Solution:

Step 1: Define the temperature dependence of resistance for metals. The resistance of metals decreases with decreasing temperature because the lattice vibrations decrease, reducing the scattering of electrons.

Step 2: Define the temperature dependence of resistance for semiconductors. The resistance of semiconductors increases with decreasing temperature because fewer electrons have enough energy to jump from the valence band to the conduction band, reducing the number of charge carriers.

Step 3: Apply to copper and germanium. Copper is a metal, so its resistance decreases. Germanium is a semiconductor, so its resistance increases.

Step 4: Conclusion: The resistance of copper decreases, and the resistance of germanium increases.

Why other options are wrong:
- Option A: This is the correct answer.
- Option B: This is the opposite of what happens.
- Option C: Both do not increase.
- Option D: Both do not decrease.

Key Concepts:
- Temperature Dependence of Resistance
- Metals
- Semiconductors

Memory Aids:
- Metals get "more conductive" when cold; semiconductors get "less conductive".

Quick Tips:
- Remember the opposite temperature dependence of resistance for metals and semiconductors.

36 Medium

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Understanding +4 | -1

In Boolean algebra,

\overline {\bar A \cdot \bar B}

is equal to

View answer and explanation

Correct option: D

A + B

Option D (

A + B

) is correct because it follows from De Morgan's theorem in Boolean algebra.

Solution approach

Correct Answer + Core Reason: $A + B$ . This is a direct application of De Morgan's theorem.

Step-by-Step Solution:

Step 1: Recall De Morgan's theorem: $\overline{X \cdot Y} = \overline{X} + \overline{Y}$

Step 2: Apply De Morgan's theorem to the given expression: $\overline{\overline{A} \cdot \overline{B}} = \overline{\overline{A}} + \overline{\overline{B}}$

Step 3: Simplify the expression: $\overline{\overline{A}} = A$ and $\overline{\overline{B}} = B$

Step 4: Therefore, $\overline{\overline{A} \cdot \overline{B}} = A + B$

Why other options are wrong:
- Option A: $\overline{A} \cdot \overline{B}$ is incorrect. It's the original expression without the outer complement.
- Option B: $\overline{A} + \overline{B}$ is incorrect. It's the complement of A AND B, not the complement of (complement of A AND complement of B).
- Option C: $A \cdot B$ is incorrect. It's the AND of A and B, not the correct application of De Morgan's theorem.
- Option D: $A + B$ is the correct answer.

Key Concepts:
- Boolean algebra
- De Morgan's theorem
- Complement operation

Memory Aids:
- Remember De Morgan's theorem: $\overline{X \cdot Y} = \overline{X} + \overline{Y}$ and $\overline{X + Y} = \overline{X} \cdot \overline{Y}$

Quick Tips:
- Always apply De Morgan's theorem carefully, remembering to complement each term and change the operator.

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Analyzing +4 | -1

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R

Assertion A: Photodiodes are used in forward bias usually for measuring the light intensity.

Reason R: For a p-n junction diode, at applied voltage

V

the current in the forward bias is more than the current in the reverse bias for

\left|V_{z}\right|> \pm V \geq

\left|V_{0}\right|

where

V_{0}

is the threshold voltage and

V_{z}

is the breakdown voltage.

In the light of the above statements, choose the correct answer from the options given below

View answer and explanation

Correct option: A

A is false but R is true

Option A (A is false but R is true) is correct because photodiodes are used in reverse bias, and the statement about forward and reverse bias current is correct.

Solution approach

Correct Answer + Core Reason: A is false but R is true. Photodiodes are used in reverse bias, and the statement about forward and reverse bias current is correct.

Step-by-Step Solution:

Step 1: Analyze Assertion A: Photodiodes are used in reverse bias, not forward bias, for measuring light intensity. In reverse bias, the small leakage current due to minority carriers is highly sensitive to light.

Step 2: Analyze Reason R: The statement about the current in forward bias being more than the current in reverse bias for a p-n junction diode is generally true.

Step 3: Determine the correctness of the statements: Assertion A is false, and Reason R is true.

Final Answer: A is false but R is true

Why other options are wrong:
- Option B: Assertion A is false.
- Option C: Assertion A is false.
- Option D: Assertion A is false.

Key Concepts:
- Photodiode
- Reverse bias
- Forward bias
- p-n junction diode

Memory Aids:
- Photodiodes operate in reverse bias.

Quick Tips:
- Remember that photodiodes are used in reverse bias to detect light.

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Understanding +4 | -1

If

{D_e}

,

{D_b}

and

{D_c}

are the doping levels of emitter, base and collector respectively of a transistor, then

View answer and explanation

Correct option: D

{D_e}

>

{D_c}

>

{D_b}

Option D (

{D_e}

>

{D_c}

>

{D_b}

) is correct because the emitter is most heavily doped to supply the majority carriers, the collector is moderately doped to collect these carriers, and the base is lightly doped to minimize recombination.

Solution approach

Correct Answer + Core Reason: ${D_e}$ > ${D_c}$ > ${D_b}$ . This order ensures proper transistor operation.

Step-by-Step Solution:

Step 1: Recall the function of each region in a transistor: emitter, base, and collector.

Step 2: Explain the emitter's role. The emitter is responsible for injecting a large number of charge carriers into the base.

Step 3: Explain the base's role. The base is a thin, lightly doped region that controls the flow of charge carriers from the emitter to the collector.

Step 4: Explain the collector's role. The collector collects the charge carriers that pass through the base.

Step 5: Relate doping levels to function. The emitter needs to be heavily doped to inject a large number of carriers, the base needs to be lightly doped to minimize recombination, and the collector needs moderate doping to efficiently collect the carriers.

Why other options are wrong:
- Option A: ${D_e}$ = ${D_b}$ = ${D_c}$ : This would not allow for proper transistor action.
- Option B: ${D_e}$ < ${D_b}$ = ${D_c}$ : This would not allow for efficient injection of carriers from the emitter.
- Option C: ${D_e}$ > ${D_b}$ > ${D_c}$ : While the emitter is most heavily doped, the collector is more doped than the base.

Key Concepts:
- Transistor structure
- Doping levels
- Emitter, base, and collector functions

Memory Aids:
- Emitter is most doped, base is least doped.

Quick Tips:
- Remember the doping order: Emitter > Collector > Base.

39 Medium

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Remembering +4 | -1

The saturation current in a diode valve is governed by

View answer and explanation

Correct option: C

Richardson’s law

Option C (Richardson’s law) is correct because Richardson's law describes the relationship between the thermionic emission current density and the temperature of the cathode in a vacuum tube.

Solution approach

Correct Answer + Core Reason: Richardson’s law. This law governs thermionic emission, which determines saturation current.

Step-by-Step Solution:

Step 1: Define saturation current. Saturation current is the maximum current that can flow in a diode valve when all emitted electrons are collected.

Step 2: Recall Richardson's law. Richardson's law describes the thermionic emission of electrons from a heated surface.

Step 3: State Richardson's equation: $J = A T^2 e^{-\frac{\phi}{kT}}$ , where J is the current density, A is Richardson's constant, T is the temperature, $\phi$ is the work function, and k is Boltzmann's constant.

Step 4: Explain the relevance. The saturation current is directly related to the current density described by Richardson's law.

Why other options are wrong:
- Option A: Child’s law: Child's law describes the space-charge-limited current in a vacuum tube, not the saturation current.
- Option B: Len’s law: Lenz's law relates to induced EMF and magnetic flux, not diode current.
- Option D: Ampere’s law: Ampere's law relates magnetic field to current, not diode current.

Key Concepts:
- Thermionic emission
- Saturation current
- Richardson's law

Memory Aids:
- Richardson's law deals with thermionic emission.

Quick Tips:
- Remember that Richardson's law is specifically for thermionic emission and saturation current in vacuum tubes.

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Understanding +4 | -1

The mobility of free electrons is greater than that of free holes because they

View answer and explanation

Correct option: B

Require low energy to continue their motion

Option B (Require low energy to continue their motion) is correct because free electrons have lower effective mass and require less energy to move through the crystal lattice compared to holes.

Solution approach

Correct Answer + Core Reason: Require low energy to continue their motion. Electrons have lower effective mass.

Step-by-Step Solution:

Step 1: Define mobility. Mobility is the measure of how easily charge carriers move through a material under the influence of an electric field.

Step 2: Explain electron and hole movement. Electrons move in the conduction band, while holes move in the valence band.

Step 3: Compare effective mass. Electrons generally have a lower effective mass than holes in most semiconductors.

Step 4: Relate effective mass to mobility. Lower effective mass means electrons require less energy to change their momentum and can move more easily through the crystal lattice.

Step 5: Conclusion: The mobility of free electrons is greater than that of free holes because they require low energy to continue their motion due to their lower effective mass.

Why other options are wrong:
- Option A: The sign of the charge does not directly determine mobility.
- Option B: This is the correct explanation.
- Option C: Electrons are lighter than the atoms, but the effective mass is what matters here, and electrons have lower effective mass than holes.
- Option D: Mutual collisions are not the primary reason for the difference in mobility.

Key Concepts:
- Mobility
- Effective Mass
- Electrons and Holes
- Semiconductor Physics

Memory Aids:
- Electrons are "lighter" and move more easily.

Quick Tips:
- Remember that mobility is inversely proportional to effective mass.

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Application +4 | -1

Two diodes have resistance 20

\mathrm{\Omega }

and is centretapped with rms secondary voltage from centre tap to each end of secondary 50 V. If external resistance is 980

\mathrm{\Omega }

. What is mean load?

View answer and explanation

Correct option: B

45 mA

Option B (45 mA) is correct because it is the approximate mean load current calculated using the given diode resistance, external resistance, and secondary voltage.

Solution approach

Correct Answer + Core Reason: 45 mA. This is the approximate mean load current.

Step-by-Step Solution:

Step 1: Calculate the peak current ( $I_p$ ).
- $I_p = \frac{V_m}{R_f + R_L} = \frac{50 V}{20 \Omega + 980 \Omega} = \frac{50 V}{1000 \Omega} = 0.05 A = 50 mA$

Step 2: Calculate the mean load current ( $I_{dc}$ ). For a center-tapped full-wave rectifier, $I_{dc} = \frac{2I_p}{\pi}$
- $I_{dc} = \frac{2 \times 50 mA}{\pi} \approx \frac{100 mA}{3.14} \approx 31.8 mA$
- However, the question states that the diodes have a resistance of 20 ohms *each*. This means the total resistance in the loop is $20 + 20 + 980 = 1020 \Omega$ .
- $I_p = \frac{50}{1020} = 0.049 A = 49 mA$
- $I_{dc} = \frac{2 \times 49}{\pi} = 31.2 mA$ . None of the answers are close.
- Let's assume the 20 ohms is *per diode*. Then the total resistance is $20+980 = 1000 \Omega$ . $I_p = 50/1000 = 0.05A$ . The average current is $2I_p/\pi = 2(0.05)/\pi = 0.0318 A = 31.8 mA$ . Still not close.
- The closest answer is 45mA, so let's assume the question meant half-wave rectification. Then $I_{dc} = I_p/\pi = 50mA/\pi = 15.9mA$ . This is also not close.
- If we assume the diode resistance is negligible, then $I_p = 50/980 = 0.051 A$ . Then $I_{dc} = 2I_p/\pi = 2(0.051)/\pi = 0.0324 A = 32.4 mA$ . Still not close.
- Let's assume the question is asking for RMS load current. $I_{rms} = I_p/\sqrt{2} = 50mA/\sqrt{2} = 35.3mA$ . Still not close.
- The question is flawed. The closest answer is 45mA, but the calculation does not support this answer.
- Let's assume the question is asking for the *peak* current. Then $I_p = 50/(20+980) = 50/1000 = 0.05 A = 50 mA$ . This is still not an option.
- Let's assume they meant *one* diode with 20 ohms. Then $I_p = 50/1000 = 0.05 A = 50 mA$ . $I_{dc} = 2*0.05/\pi = 0.0318 A = 31.8 mA$ .
- If we assume that the question meant that the *total* diode resistance is 20 ohms, then $I_p = 50/(20+980) = 0.05 A$ . The average current is $2I_p/\pi = 2(0.05)/\pi = 0.0318 A = 31.8 mA$ . Still not close to 45mA.
- Let's assume that the question meant to say that the *peak* secondary voltage is 50V, not RMS. Then $V_m = 50 \sqrt{2} = 70.7 V$ . Then $I_p = 70.7/1000 = 0.0707 A = 70.7 mA$ . Then $I_{dc} = 2(0.0707)/\pi = 0.045 A = 45 mA$ . This works!

Why other options are wrong:
- Option A: 0.05 A is incorrect. It's the peak current, not the mean load current.
- Option B: 45 mA is the correct answer (assuming the 50V is RMS and we need to convert to peak voltage).
- Option C: 0.25 A is incorrect. It's a calculation error.
- Option D: 25 mA is incorrect. It's a calculation error.

Key Concepts:
- Full-wave rectifier
- Center-tapped transformer
- Mean load current
- Peak current

Memory Aids:
- Remember the formula: $I_{dc} = \frac{2I_p}{\pi}$ for a center-tapped full-wave rectifier.

Quick Tips:
- Always calculate the peak current first, then calculate the mean load current.

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Remembering +4 | -1

The temperature coefficient of resistance of a conductor is

View answer and explanation

Correct option: A

Positive always

Option A (Positive always) is correct because the resistance of a conductor generally increases with temperature.

Solution approach

Correct Answer + Core Reason: Positive always. The resistance of conductors typically increases with increasing temperature.

Step-by-Step Solution:

Step 1: Define temperature coefficient of resistance: The temperature coefficient of resistance, often denoted by $\alpha$ , describes the change in electrical resistance of a material per degree Celsius (or Kelvin) change in temperature.

Step 2: Consider the behavior of conductors: In most metallic conductors, as temperature increases, the atoms vibrate more vigorously, hindering the flow of electrons. This leads to an increase in resistance.

Step 3: Mathematical representation: The relationship between resistance and temperature is often approximated by the equation: $R = R_0[1 + \alpha(T - T_0)]$

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Analyzing +4 | -1

Assertion : Photodiode & photovoltaic cell are based on the same principle.

Reason : Both use same method of operations to work.

View answer and explanation

Correct option: C

If Assertions is True but the reason is False

Option C is correct. While both photodiode and photovoltaic cells use semiconductors and p-n junctions, they operate on different principles. Photodiodes are typically reverse biased and detect light, while photovoltaic cells generate voltage from light.

Solution approach

Correct Answer + Core Reason: If Assertions is True but the reason is False. Photodiodes and photovoltaic cells are based on similar semiconductor physics, but their operation methods differ.

Step-by-Step Solution:

Step 1: Understand the principle of operation of a photodiode. A photodiode is a semiconductor diode that operates in reverse bias. When light (photons) falls on the depletion region of the diode, it generates electron-hole pairs. These carriers are then swept across the junction due to the electric field, resulting in a current flow. The current is proportional to the intensity of the incident light.

Step 2: Understand the principle of operation of a photovoltaic cell (solar cell). A photovoltaic cell also uses a p-n junction. When light falls on the junction, electron-hole pairs are generated. However, unlike a photodiode, a photovoltaic cell is designed to generate a voltage. The separation of electron-hole pairs creates a potential difference, which can be used to drive an external circuit.

Step 3: Compare the two devices. Both devices utilize the photoelectric effect in a semiconductor p-n junction. However, a photodiode is typically used to *detect* light (measuring current), while a photovoltaic cell is used to *generate* electricity (producing voltage).

Step 4: Analyze the Assertion. The assertion that both devices are based on the same principle (photoelectric effect in a p-n junction) is true.

Step 5: Analyze the Reason. The reason states that both use the same method of operations to work. This is false. Photodiodes are usually reverse-biased and measure current, while photovoltaic cells generate voltage without external bias.

Why other options are wrong:
- Option A: The reason is false.
- Option B: The reason is false.
- Option D: The assertion is true.

Key Concepts:
- Photodiode operation
- Photovoltaic cell operation
- p-n junction
- Photoelectric effect

Memory Aids:
- Photodiode: Detects light (current)
- Photovoltaic cell: Generates voltage

Quick Tips:
- Understand the difference between detecting light and generating electricity using semiconductors.

44 Medium

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Understanding +4 | -1

Would there be any advantage to adding

n

-type or

p

-type impurities to copper

View answer and explanation

Correct option: B

No

No, because copper is already a good conductor, and adding impurities would decrease its conductivity.

Solution approach

Correct Answer + Core Reason: No. Copper is already an excellent conductor, and adding impurities would decrease its conductivity.

Step-by-Step Solution:

Step 1: Recall the effect of impurities on conductivity in metals.

Step 2: Copper is a good conductor due to its free electrons.

Step 3: Adding n-type or p-type impurities would introduce scattering centers.

Step 4: These scattering centers would reduce the mean free path of electrons and decrease conductivity.

Step 5: Conclusion: There is no advantage to adding impurities to copper.

Why other options are wrong:
- Option A: Adding impurities would decrease conductivity, not increase it.
- Option C: There is no scenario where adding impurities would improve copper's conductivity.
- Option D: The information is sufficient to determine that adding impurities would be detrimental.

Key Concepts:
- Conductivity in metals
- Impurities and scattering
- Mean free path

Memory Aids:
- Pure metals are best conductors; impurities always reduce conductivity.

Quick Tips:
- Remember that doping is primarily used to control conductivity in semiconductors, not metals.

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Remembering +4 | -1

In a P-type semiconductor, the acceptor impurity produces an energy level

View answer and explanation

Correct option: D

Just above the valence band

Option D (Just above the valence band) is correct because acceptor impurities create energy levels slightly above the valence band, allowing electrons to easily jump into them, creating holes.

Solution approach

Correct Answer + Core Reason: Just above the valence band. Acceptor impurities create energy levels slightly above the valence band.

Step-by-Step Solution:

Step 1: Define acceptor impurities. These are impurities added to a semiconductor to create a P-type semiconductor. They accept electrons, creating holes.

Step 2: Explain the energy levels created by acceptor impurities. Acceptor impurities create energy levels just above the valence band. Electrons from the valence band can easily jump into these levels, creating holes in the valence band.

Step 3: Conclusion: In a P-type semiconductor, the acceptor impurity produces an energy level just above the valence band.

Why other options are wrong:
- Option A (Just below the valence band): This is incorrect. Acceptor levels are above the valence band.
- Option B (Just above the conduction band): This is incorrect. Donor levels are below the conduction band.
- Option C (Just below the conduction band): This is incorrect. Donor levels are below the conduction band.
- Option D (Just above the valence band): This is the correct option.

Key Concepts:
- P-type semiconductor
- Acceptor impurities
- Valence band
- Energy levels

Memory Aids:
- Remember that acceptor levels are "a"bove the valence band.

Quick Tips:
- Understand the energy band diagrams of semiconductors.

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Application +4 | -1

A common emitter amplifier is designed with

NPN

transistor

\left( {\alpha = 0.99} \right)

. The input impedance is

1\;K\mathrm{\Omega }

and load is

10\;K\mathrm{\Omega }

. The voltage gain will be

View answer and explanation

Correct option: C

990

Option C (990) is correct because the voltage gain of a common emitter amplifier is approximately equal to the product of beta and the ratio of load resistance to input impedance.

Solution approach

Correct Answer + Core Reason: 990. Voltage gain is approximately equal to the product of beta and the ratio of load resistance to input impedance.

Step-by-Step Solution:

Step 1: Calculate β from α.
- $\beta = \frac{\alpha}{{1 - \alpha}} = \frac{0.99}{{1 - 0.99}} = \frac{0.99}{0.01} = 99$

Step 2: Calculate the voltage gain ( $A_v$ ).
- $A_v = \beta \times \frac{R_L}{R_i} = 99 \times \frac{10 k\Omega}{1 k\Omega} = 99 \times 10 = 990$

Why other options are wrong:
- Option A: 9.9 is incorrect. It might be a calculation error.
- Option B: 99 is incorrect. It might be a calculation error.
- Option C: 990 is the correct answer.
- Option D: 9900 is incorrect. It might be a calculation error.

Key Concepts:
- Common emitter amplifier
- Voltage gain
- BJT parameters (α, β)

Memory Aids:
- Remember the formula: $A_v = \beta \times \frac{R_L}{R_i}$

Quick Tips:
- Always calculate β from α first, if α is given.

47 Medium

47 of 85Question permalink

Remembering +4 | -1

In Colpitt oscillator the feedback network consists of

View answer and explanation

Correct option: B

Two capacitors and an inductor

Option B (Two capacitors and an inductor) is correct because a Colpitts oscillator uses a feedback network consisting of two capacitors and one inductor to create the necessary phase shift for oscillation.

Solution approach

Correct Answer + Core Reason: Two capacitors and an inductor. This is the defining characteristic of a Colpitts oscillator.

Step-by-Step Solution:

Step 1: Recall the basic structure of an oscillator. An oscillator consists of an amplifier and a feedback network.

Step 2: Describe the Colpitts oscillator feedback network. The Colpitts oscillator uses a tank circuit consisting of two capacitors in series and an inductor in parallel with the capacitor combination.

Step 3: Explain the function of the feedback network. The tank circuit provides the necessary positive feedback and determines the oscillation frequency.

Why other options are wrong:
- Option A: Two inductors and a capacitor: This describes a Hartley oscillator.
- Option C: Three pairs of $RC$ circuit: This is not a typical configuration for an oscillator.
- Option D: Three pairs of $RL$ circuit: This is not a typical configuration for an oscillator.

Key Concepts:
- Oscillators
- Feedback networks
- Colpitts oscillator
- Tank circuit

Memory Aids:
- Colpitts = Capacitors

Quick Tips:
- Remember the difference between Colpitts and Hartley oscillators.

48 Medium

48 of 85Question permalink

Understanding +4 | -1

The output of a NAND gate is 0

View answer and explanation

Correct option: C

If both inputs are 1

Option C (If both inputs are 1) is correct because a NAND gate outputs 0 only when both inputs are 1.

Solution approach

Correct Answer + Core Reason: If both inputs are 1. NAND gate gives output 0 only when both inputs are 1.

Step-by-Step Solution:

Step 1: Recall the truth table for a NAND gate.

Step 2: The NAND gate is the complement of the AND gate. The AND gate outputs 1 only when both inputs are 1. Therefore, the NAND gate outputs 0 only when both inputs are 1.

Step 3: Consider the other input combinations:
- If both inputs are 0, the output is 1.
- If one input is 0 and the other is 1, the output is 1.

Why other options are wrong:
- Option A: If both inputs are 0, the output is 1, not 0.
- Option B: If one input is 0 and the other input is 1, the output is 1, not 0.
- Option C: If both inputs are 1, the output is 0, which is the correct answer.
- Option D: This option is incorrect because it includes cases where the output is 1.

Key Concepts:
- Logic gates (NAND, AND)
- Truth tables

Memory Aids:
- NAND: NOT AND (complement of AND)

Quick Tips:
- Memorize the truth tables for common logic gates.

49 Medium

49 of 85Question permalink

Understanding +4 | -1

A LED (light emitting diode) in its proper Operation is

View answer and explanation

Correct option: B

A heavily doped p-n junction in forward biased

Option B (A heavily doped p-n junction in forward biased) is correct because LEDs require a heavily doped p-n junction to enhance radiative recombination and are forward biased to allow current flow and light emission.

Solution approach

Correct Answer + Core Reason: A heavily doped p-n junction in forward biased. This configuration allows for efficient light emission.

Step-by-Step Solution:

Step 1: Describe the structure of an LED. An LED is a specially designed p-n junction diode.

Step 2: Explain the doping levels. LEDs are heavily doped to increase the concentration of charge carriers.

Step 3: Explain the biasing. LEDs are forward biased to allow current to flow.

Step 4: Describe the light emission process. When electrons and holes recombine in the depletion region, energy is released in the form of photons (light).

Why other options are wrong:
- Option A: A heavily doped p-n junction in reverse biased: Reverse bias does not allow significant current flow, so no light is emitted.
- Option C: A lightly doped p-n junction in reverse biased: Light doping reduces the efficiency of recombination.
- Option D: Lightly doped p-n junction in no bias: No bias means no current flow, so no light is emitted.

Key Concepts:
- LED structure
- Doping levels
- Forward bias
- Light emission
- Recombination

Memory Aids:
- LED = Forward, Heavy

Quick Tips:
- Remember that LEDs require forward bias and heavy doping for efficient light emission.

50 Medium

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Analyzing +4 | -1

Assertion: A P-N photodiode is made from a semiconductor for which

\mathrm{Eg}=2.8 \mathrm{eV}

. This photodiode will not detect the wavelength of

6000 \mathrm{~nm}

.

Reason: A PN photodiode detects wavelength I if

\frac{h c}{\lambda} > E_g

View answer and explanation

Correct option: B

Both Assertion and Reason are true, but Reason is not the correct explanation for Assertion

Option B is correct because both the assertion and reason are true, but the reason doesn't fully explain the assertion. The photodiode won't detect the wavelength because the photon energy is less than the band gap energy, and the reason correctly states the condition for detection.

Solution approach

Correct Answer + Core Reason: Both Assertion and Reason are true, but Reason is not the correct explanation for Assertion. The assertion is true because the photon energy is less than the band gap, and the reason correctly states the condition for detection.

Step-by-Step Solution:

Step 1: Calculate the energy corresponding to the given wavelength using the formula $E = \frac{{hc}}{\lambda}$ .

Step 2: Convert the wavelength from nm to meters: $\lambda = 6000 \text{{ nm}} = 6000 \times 10^{-9} \text{{ m}}$ .

Step 3: Substitute the values of $h$ , $c$ , and $\lambda$ into the energy formula: $E = \frac{{(6.626 \times 10^{-34} \text{{ Js}})(3 \times 10^8 \text{{ m/s}})}}{{6000 \times 10^{-9} \text{{ m}}}} \approx 3.313 \times 10^{-20} \text{{ J}}$ .

Step 4: Convert the energy from Joules to eV: $E = \frac{{3.313 \times 10^{-20} \text{{ J}}}}{{1.602 \times 10^{-19} \text{{ C}}}} \approx 0.207 \text{{ eV}}$ .

Step 5: Compare the calculated energy (0.207 eV) with the band gap energy (2.8 eV). Since 0.207 eV < 2.8 eV, the photodiode will not detect the wavelength of 6000 nm, making the assertion true.

Step 6: The reason states that a photodiode detects wavelength $\lambda$ if $\frac{{hc}}{\lambda} > E_g$ . This is a correct statement.

Step 7: While both the assertion and reason are true, the reason doesn't fully explain *why* the photodiode is made from a semiconductor with Eg = 2.8 eV. The band gap is chosen to detect a specific range of wavelengths, but the reason only provides the general condition for detection.

Why other options are wrong:
- Option A: The reason is not the *correct* explanation for the assertion; it's a related but not fully explanatory statement.
- Option C: The reason is true.
- Option D: The assertion is true.

Key Concepts:
- Energy of a photon: $E = \frac{{hc}}{\lambda}$
- Band gap energy of a semiconductor
- Photodiode operation

Memory Aids:
- Remember the relationship between energy, wavelength, and frequency: $E = hf = \frac{{hc}}{\lambda}$

Quick Tips:
- Always convert units to SI units before calculations.
- Remember the values of Planck's constant ( $h$ ) and the speed of light ( $c$ ).

51 Medium

51 of 85Question permalink

Understanding +4 | -1

In a transistor the collector current is always less than the emitter current because

View answer and explanation

Correct option: B

A few electrons are lost in the base and only remaining ones reach the collector

Option B (A few electrons are lost in the base and only remaining ones reach the collector) is correct because some of the electrons injected from the emitter recombine with holes in the base region, reducing the number of electrons that reach the collector.

Solution approach

Correct Answer + Core Reason: A few electrons are lost in the base and only remaining ones reach the collector. Recombination in the base reduces collector current.

Step-by-Step Solution:

Step 1: Describe the current flow in a transistor. Electrons are injected from the emitter into the base in an NPN transistor.

Step 2: Explain recombination in the base. Some of these electrons recombine with holes in the base region.

Step 3: Explain the effect on collector current. The electrons that do not recombine in the base reach the collector, contributing to the collector current.

Step 4: Conclude that collector current is less than emitter current. Because of recombination in the base, the collector current is always less than the emitter current.

Why other options are wrong:
- Option A: Collector side is reverse biased and the emitter side is forward biased: This is the normal operating condition, but it doesn't directly explain why $I_C < I_E$ .
- Option C: Collector being reverse biased, attracts less electrons: The reverse bias helps collect the electrons that reach the collector.
- Option D: Collector side is forward biased and emitter side is reverse biased: This is not the normal operating condition for a transistor.

Key Concepts:
- Transistor operation
- Current flow
- Recombination
- Emitter current, base current, collector current

Memory Aids:
- Recombination reduces current.

Quick Tips:
- Remember that $I_E = I_B + I_C$ , and $I_B$ is always greater than zero.

52 Medium

52 of 85Question permalink

Understanding +4 | -1

Number of secondary electrons emitted per number of primary electrons depends on

View answer and explanation

Correct option: C

Intensity

Intensity is correct because the number of secondary electrons emitted depends on the number of primary electrons, which is related to the intensity.

Solution approach

Correct Answer + Core Reason: Intensity. The number of secondary electrons emitted per primary electron depends on the intensity of the primary electrons.

Step-by-Step Solution:

Step 1: Define secondary emission.

Step 2: Secondary emission is the emission of electrons from a material due to bombardment by other energetic particles (primary electrons).

Step 3: The number of secondary electrons emitted depends on the energy and number of primary electrons.

Step 4: Intensity is related to the number of primary electrons.

Step 5: Conclusion: The number of secondary electrons depends on the intensity.

Why other options are wrong:
- Option A: The material of the target affects the *efficiency* of secondary emission, but not directly the number of secondary electrons per primary electron. The intensity of the primary beam is the dominant factor.
- Option B: The frequency of primary electrons is related to their energy, but the number of secondary electrons emitted is more directly related to the number of primary electrons (intensity).
- Option D: Intensity is a factor.

Key Concepts:
- Secondary emission
- Intensity
- Primary electrons

Memory Aids:
- More intense primary beam = more secondary electrons.

Quick Tips:
- Remember that intensity is related to the number of particles.

53 Medium

53 of 85Question permalink

Remembering +4 | -1

In a transistor the base is

View answer and explanation

Correct option: A

An insulator

Option A (An insulator) is correct because the base of a transistor is very thin and lightly doped, effectively acting as an insulator to control the current flow.

Solution approach

Correct Answer + Core Reason: An insulator. The base region is very thin and lightly doped, limiting current flow.

Step-by-Step Solution:

Step 1: Understand the structure of a transistor. A transistor consists of three regions: emitter, base, and collector.

Step 2: Consider the doping levels and width of each region. The base is lightly doped and very thin compared to the emitter and collector.

Step 3: Relate doping level to conductivity. Light doping implies low conductivity, approaching insulating behavior.

Step 4: Explain the function of the base. The base controls the current flow between the emitter and collector. A small change in base current can cause a large change in collector current.

Step 5: Conclude that the base acts as a control element, not a good conductor.

Why other options are wrong:
- Option B: A conductor of low resistance: The base is lightly doped, so it does not have low resistance.
- Option C: A conductor of high resistance: While the base has higher resistance than the emitter or collector, it's not considered a high-resistance conductor in the typical sense.
- Option D: An extrinsic semiconductor: While the base is made of a semiconductor material, its primary function in the transistor circuit is not to act as a typical semiconductor but as a control element.

Key Concepts:
- Transistor structure (emitter, base, collector)
- Doping levels in semiconductors
- Conductivity and resistance
- Transistor operation

Memory Aids:
- Base is thin and lightly doped.

Quick Tips:
- Remember the relative doping levels: Emitter is heavily doped, base is lightly doped, and collector is moderately doped.

54 Medium

54 of 85Question permalink

Remembering +4 | -1

Which of the following materials in non crystalline

View answer and explanation

Correct option: C

Wood

Wood is correct because it lacks long-range order and is therefore non-crystalline.

Solution approach

Correct Answer + Core Reason: Wood. Wood lacks a repeating, long-range order of atoms, making it non-crystalline.

Step-by-Step Solution:

Step 1: Define crystalline and non-crystalline materials.

Step 2: Crystalline materials have a long-range, repeating order of atoms.

Step 3: Non-crystalline (amorphous) materials lack this long-range order.

Step 4: Consider the given options: Copper, Sodium chloride, and Diamond are crystalline. Wood is not.

Step 5: Conclusion: Wood is non-crystalline.

Why other options are wrong:
- Option A: Copper is a metal with a crystalline structure.
- Option B: Sodium chloride is an ionic compound with a crystalline structure.
- Option C: Diamond is a form of carbon with a crystalline structure.
- Option D: Diamond is a crystalline allotrope of carbon.

Key Concepts:
- Crystalline vs. Non-crystalline materials
- Atomic structure

Memory Aids:
- Think of crystals as having a regular, repeating pattern.

Quick Tips:
- Recognize common crystalline materials like metals and ionic compounds.

55 Medium

55 of 85Question permalink

Remembering +4 | -1

Name of a

p

-

n

junction, which can be used as the regulator, is

View answer and explanation

Correct option: A

Zener diode

Option A (Zener diode) is correct because Zener diodes are specifically designed to operate in the reverse breakdown region and maintain a constant voltage.

Solution approach

Correct Answer + Core Reason: Zener diode. Zener diodes are designed for voltage regulation.

Step-by-Step Solution:

Step 1: Define voltage regulators. Voltage regulators maintain a constant output voltage despite variations in input voltage or load current.

Step 2: Consider the properties of Zener diodes. Zener diodes have a sharp reverse breakdown voltage and can operate in the breakdown region without damage.

Step 3: Consider the properties of Tunnel and Gunn diodes. Tunnel diodes exhibit negative resistance and are used in high-frequency oscillators. Gunn diodes are used for microwave signal generation.

Step 4: Conclusion: Zener diodes are used as voltage regulators.

Why other options are wrong:
- Option A: Zener diodes are designed for voltage regulation in the reverse breakdown region.
- Option B: Tunnel diodes are used in high-frequency oscillators.
- Option C: Gunn diodes are used for microwave signal generation.
- Option D: Zener diode is the correct answer.

Key Concepts:
- Zener Diode Characteristics
- Voltage Regulation
- Tunnel Diode
- Gunn Diode

Memory Aids:
- Zener = Regulator (Think of Zener as being in control, like a regulator).

Quick Tips:
- Remember the primary applications of different types of diodes.

56 Medium

56 of 85Question permalink

Remembering +4 | -1

In silicon when phosphorus is doped ……. is formed

View answer and explanation

Correct option: B

n

-type semiconductor

Option B (

n

-type semiconductor) is correct because doping silicon with phosphorus, a pentavalent element, creates an n-type semiconductor.

Solution approach

Correct Answer + Core Reason: $n$ -type semiconductor. Phosphorus is a pentavalent impurity.

Step-by-Step Solution:

Step 1: Silicon is a group IV element with four valence electrons.

Step 2: Phosphorus is a group V element with five valence electrons.

Step 3: When phosphorus is added to silicon (doping), four of phosphorus's valence electrons form covalent bonds with silicon atoms, leaving one extra electron.

Step 4: This extra electron is free to move and contributes to electrical conductivity. Since the extra charge carriers are negative (electrons), the resulting semiconductor is called an n-type semiconductor.

Why other options are wrong:
- Option A: A p-type semiconductor is formed by doping with a trivalent element (like boron).
- Option C: A p-n junction is formed when a p-type and an n-type semiconductor are joined together.
- Option D: One of the other options is correct.

Key Concepts:
- N-type semiconductors
- Doping
- Pentavalent impurities
- Silicon

Memory Aids:
- N stands for negative, indicating that electrons are the majority carriers in n-type semiconductors.
- Remember that pentavalent impurities create n-type semiconductors.

Quick Tips:
- Know the difference between n-type and p-type semiconductors and the types of impurities used to create them.

57 Medium

57 of 85Question permalink

Application +4 | -1

If the slope of the characteristics curve of a vacuum tube diode corresponding to some point is

{10^{ - 3}}\mathrm{mA}{\mathrm{V}^{ - 1}}

, then the plate resistance of diode and its nature are respectively

View answer and explanation

Correct option: B

1000\mathrm{\;k\Omega }

, dynamic

Option B (

1000\mathrm{\\;k\Omega }

, dynamic) is correct because the plate resistance is the inverse of the slope of the characteristic curve, and it represents a dynamic resistance.

Solution approach

Correct Answer + Core Reason: $1000\mathrm{\\;k\Omega }$ , dynamic. Plate resistance is the inverse of the slope of the characteristic curve, representing dynamic resistance.

Step-by-Step Solution:

Step 1: Identify the given information: Slope $= {10^{ - 3}}\mathrm{mA}{\mathrm{V}^{ - 1}} = {10^{ - 6}}\mathrm{A}{\mathrm{V}^{ - 1}}$

Step 2: Calculate the plate resistance: $r_p = \frac{1}{{{\text{{slope}}}}} = \frac{1}{{{{10}^{ - 6}}\mathrm{A}{\mathrm{V}^{ - 1}}}} = {10^6}\Omega = 1000\mathrm{\\;k\Omega }$

Step 3: Determine the nature of the resistance: Since it's derived from the slope of a curve, it's a dynamic resistance.

Why other options are wrong:
- **Option A ( $100\mathrm{\\;k\Omega }$ , static): Incorrect calculation of resistance and incorrect nature.
- Option C ( $100\mathrm{\\;k\Omega }$ , dynamic): Incorrect calculation of resistance.
- Option D ( $1000\mathrm{\\;k\Omega }$ , static): Correct calculation of resistance but incorrect nature.

Key Concepts:
- Plate resistance of a vacuum tube diode
- Dynamic resistance

Memory Aids:
- Plate resistance is the inverse of the slope.

Quick Tips:**
- Remember the definition of plate resistance.

58 Medium

58 of 85Question permalink

Remembering +4 | -1

Which of the following has negative temperature coefficient of resistance

View answer and explanation

Correct option: D

Germanium

Option D (Germanium) is correct because semiconductors like Germanium have a negative temperature coefficient of resistance, meaning their resistance decreases with increasing temperature.

Solution approach

Correct Answer + Core Reason: Germanium. Semiconductors have a negative temperature coefficient of resistance.

Step-by-Step Solution:

Step 1: The temperature coefficient of resistance describes how the resistance of a material changes with temperature.

Step 2: Materials with a positive temperature coefficient of resistance (like most metals) have increasing resistance with increasing temperature.

Step 3: Materials with a negative temperature coefficient of resistance have decreasing resistance with increasing temperature.

Step 4: Semiconductors (like germanium and silicon) have a negative temperature coefficient of resistance because increasing temperature increases the number of free charge carriers, which reduces the resistance.

Why other options are wrong:
- Option A: Copper is a metal and has a positive temperature coefficient of resistance.
- Option B: Aluminium is a metal and has a positive temperature coefficient of resistance.
- Option C: Iron is a metal and has a positive temperature coefficient of resistance.

Key Concepts:
- Temperature coefficient of resistance
- Semiconductors
- Metals

Memory Aids:
- Remember that semiconductors have a negative temperature coefficient of resistance.
- Metals generally have a positive temperature coefficient of resistance.

Quick Tips:
- Know the difference in temperature coefficient of resistance between metals and semiconductors.

59 Medium

59 of 85Question permalink

Remembering +4 | -1

In an intrinsic semiconductor, the Fermi level is

View answer and explanation

Correct option: B

Equidistance from conduction band and valency band

Option B (Equidistance from conduction band and valency band) is correct because in an intrinsic semiconductor, the Fermi level lies in the middle of the band gap.

Solution approach

Correct Answer + Core Reason: Equidistance from conduction band and valency band. In an intrinsic semiconductor, the Fermi level lies in the middle of the band gap.

Step-by-Step Solution:

Step 1: Define intrinsic semiconductor: An intrinsic semiconductor is a pure semiconductor without any impurities.

Step 2: Define Fermi level: The Fermi level is the energy level at which the probability of finding an electron is 50%.

Step 3: In an intrinsic semiconductor, the number of electrons in the conduction band is equal to the number of holes in the valence band. Therefore, the Fermi level lies in the middle of the band gap, equidistant from the conduction and valence bands.

Final Answer: Equidistance from conduction band and valency band

Why other options are wrong:
- Option A: Incorrect position of the Fermi level.
- Option C: Incorrect position of the Fermi level.
- Option D: Incorrect position of the Fermi level.

Key Concepts:
- Intrinsic semiconductor
- Fermi level
- Conduction band
- Valence band

Memory Aids:
- In intrinsic semiconductors, the Fermi level is in the middle of the band gap.

Quick Tips:
- Remember the position of the Fermi level in intrinsic semiconductors.

60 Medium

60 of 85Question permalink

Understanding +4 | -1

The potential in depletion layer is due to

View answer and explanation

Correct option: C

Ions

Option C (Ions) is correct because the potential in the depletion layer is due to the immobile ions left behind after the diffusion of charge carriers.

Solution approach

Correct Answer + Core Reason: Ions. The potential barrier is due to immobile ions.

Step-by-Step Solution:

Step 1: Describe the formation of the depletion layer. When a p-n junction is formed, electrons from the n-side diffuse to the p-side, and holes from the p-side diffuse to the n-side.

Step 2: Explain the role of ions. This diffusion leaves behind immobile ionized dopant atoms (ions) near the junction. These ions create an electric field and a potential difference across the depletion layer.

Step 3: Conclusion: The potential in the depletion layer is due to these immobile ions.

Why other options are wrong:
- Option A: Electrons are mobile charge carriers that diffuse, but the potential is due to the remaining ions.
- Option B: Holes are mobile charge carriers that diffuse, but the potential is due to the remaining ions.
- Option C: This is the correct answer. The potential is due to the immobile ions.
- Option D: The forbidden band is an energy range, not a source of potential.

Key Concepts:
- Depletion Layer
- P-N Junction
- Ion Formation
- Potential Barrier

Memory Aids:
- Ions are immobile and create the electric field.

Quick Tips:
- Remember that the depletion layer is depleted of mobile charge carriers but contains immobile ions.

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