CUET UG Physics Booster Test 2-Junction Formation and Physics -

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CUET UG Physics Booster Test 2-Junction Formation and Physics

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

CUET UG Physics Booster Test 2-Junction Formation and Physics

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QUESTION 1 OF 20

Incorrect statement about the wafer conversion process

Answer Review

1 Incorrect statement about the wafer conversion process

A It requires adding precisely a small quantity of pentavalent impurity to a p-Si wafer.

B The entire p-Si wafer is completely converted into an n-type semiconductor.

C The process creates a p-region and an n-region within the same continuous wafer.

D It avoids the discontinuity that would occur from physically joining two slabs.

💡 Short Explanation

Only a small region is converted into n-type. The remaining portion stays p-type. This creates a p-n junction.

✅ Correct Answer Explanation

During p-n junction fabrication, only a selected portion of the p-type silicon wafer is doped with pentavalent impurities. This creates an n-region while the rest remains p-type.

❓ Why Other Options Are Incorrect

Option A → Correct fabrication process.
Option C → Correct result of wafer conversion.
Option D → Correct advantage over joining separate slabs.

🎯 Strategy

Used

Conceptual Elimination

Application:

Identify the statement inconsistent with p-n junction fabrication.

Final Logic:

Only part of the wafer is converted to n-type.

🧠 Memory Trick

"One Wafer, Two Regions."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Formation of p-n Junction

2 Correct statements about the metallurgical junction
1. It acts as a discontinuity for flowing charge carriers if formed by physically pressing slabs together.
2. The inter-atomic crystal spacing is roughly 2 to 3 Å.
3. Physical roughness prevents continuous contact at the atomic level in physically joined slabs.
4. A highly efficient p-n junction can be formed by smoothly pressing an n-slab to a p-slab.

A (A), (B), (C)

B (B), (C), (D)

C (A), (C), (D)

D (A), (B), (D)

💡 Short Explanation

Atomic-scale continuity is essential. Physically joined slabs cannot provide a proper junction. Crystal spacing is extremely small.

✅ Correct Answer Explanation

Statements A, B, and C are correct. Physical roughness prevents perfect atomic contact, making simple pressing of p-type and n-type slabs ineffective for producing a high-quality p-n junction.

❓ Why Other Options Are Incorrect

Statement D → Physically pressing slabs does not create an efficient junction.

🎯 Strategy

Used

Statement Analysis

Application:

Evaluate each statement regarding junction formation.

Final Logic:

A, B, and C are correct.

🧠 Memory Trick

"Atomic Continuity Required."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Metallurgical Junction

3 The concentration gradient during p-n junction formation
1. is higher for free electrons in the p-type semiconductor.
2. drives holes from the p-side to the n-side.
3. drives electrons from the n-side to the p-side.
4. ceases to exist the exact moment the depletion region is fully formed.

A (A) and

B (B) and

C (C) and

D only

💡 Short Explanation

Holes diffuse from p to n. Electrons diffuse from n to p. Diffusion is driven by concentration differences.

✅ Correct Answer Explanation

The p-side contains many holes while the n-side contains many electrons. Therefore: Holes diffuse from p-side to n-side. Electrons diffuse from n-side to p-side.

❓ Why Other Options Are Incorrect

Statement A → Electron concentration is higher in n-type material.
Statement D → Concentration gradient still exists at equilibrium; it is balanced by drift.

🎯 Strategy

Used

Carrier Diffusion Analysis

Application:

Determine diffusion directions.

Final Logic:

Only B and C are correct.

🧠 Memory Trick

"Holes: p→n, Electrons: n→p."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Diffusion Process in p-n Junction

4 Match List I with List II regarding currents in a p-n junction

List I	List II
I. Initial state	P. Concentration gradient
II. Equilibrium state	Q. Diffusion current is large, drift is small
III. Diffusion current driver	R. Electric field
IV. Drift current driver	S. Net current is zero

A I-Q, II-S, III-P, IV-R

B I-S, II-Q, III-R, IV-P

C I-Q, II-S, III-R, IV-P

D I-P, II-R, III-Q, IV-S

💡 Short Explanation

Diffusion dominates initially. Net current becomes zero at equilibrium. Diffusion is driven by concentration gradient. Drift is driven by electric field.

✅ Correct Answer Explanation

Correct matching: I → Q II → S III → P IV → R

❓ Why Other Options Are Incorrect

Incorrect matching of diffusion and drift mechanisms.

🎯 Strategy

Used

Matching Concepts

Application:

Associate current mechanisms with their causes.

Final Logic:

Only Option A is correct.

🧠 Memory Trick

"Gradient → Diffusion, Field → Drift."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Diffusion and Drift Currents

5 If the strength of the electric field in the depletion region is E, and the thickness of the region is d, the potential difference V₀ is roughly modeled as

A (E × d)

B -(E × d)

C (E / d)

D (d / E)

💡 Short Explanation

For a uniform electric field: \(V=E\times d\)

✅ Correct Answer Explanation

The potential difference across a region of thickness \(d\) subjected to electric field \(E\) is approximately: \(V_{0}=Ed\)

❓ Why Other Options Are Incorrect

B → Incorrect sign convention.
C → Incorrect dimensions.
D → Incorrect dimensions.

🎯 Strategy

Used

Formula Recall

Application:

Use the electric field–potential relation.

Final Logic:

Potential difference equals field × distance.

🧠 Memory Trick

"Voltage = Field × Distance."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Barrier Potential

6 In a p-n junction under equilibrium

A the drift current and diffusion current are both inherently zero

B the drift current exceeds the diffusion current

C the drift current is equal and opposite to the diffusion current

D the drift current only exists when an external forward bias is applied

💡 Short Explanation

Diffusion and drift both exist. Their magnitudes are equal. Net current becomes zero.

✅ Correct Answer Explanation

At equilibrium: \(I_{drift}=I_{diffusion}\) but in opposite directions. Hence, the total current is zero.

❓ Why Other Options Are Incorrect

A → Both currents exist.
B → Currents are equal.
D → Drift current exists even without external bias.

🎯 Strategy

Used

Equilibrium Condition

Application:

Apply the equilibrium current condition.

Final Logic:

Opposing currents balance each other.

🧠 Memory Trick

"Drift = Diffusion at Equilibrium."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Equilibrium in p-n Junction

7 When an electron diffuses from n to p, the ionised donor left behind is ____, creating a layer of ____ charge on the n-side.

A mobile, positive

B immobile, negative

C mobile, negative

D immobile, positive

💡 Short Explanation

Donor atom loses an electron. Positive ion remains fixed in lattice.

✅ Correct Answer Explanation

After electron diffusion, donor atoms become positively charged ions. Since they are bonded within the crystal lattice, they remain immobile.

❓ Why Other Options Are Incorrect

A → Ion is not mobile.
B → Donor ions are positive.
C → Incorrect charge and mobility.

🎯 Strategy

Used

Charge Analysis

Application:

Determine donor-ion characteristics.

Final Logic:

Ionized donor = immobile positive ion.

🧠 Memory Trick

"Donor Loses Electron → Positive Ion."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Space-Charge Region

8 Space-charge region formation statements
1. Diffusing holes leave behind ionized acceptors.
2. Ionized acceptors represent a positive space-charge.
3. Ionized acceptors are immobile due to bonding with surrounding atoms.
4. The negative space-charge region forms on the p-side.

A (A), (C), (D)

B (A), (B), (C)

C (B), (C), (D)

D (A), (B), (D)

💡 Short Explanation

Hole diffusion exposes acceptor ions. Acceptor ions are negative and immobile. Negative space charge forms on p-side.

✅ Correct Answer Explanation

Statements A, C, and D are correct. Ionized acceptors possess negative charge, so statement B is incorrect.

❓ Why Other Options Are Incorrect

Statement B → Acceptor ions create negative space charge.

🎯 Strategy

Used

Space-Charge Analysis

Application:

Determine the nature of acceptor ions.

Final Logic:

A, C, and D are correct.

🧠 Memory Trick

"Acceptor Ion = Negative Fixed Charge."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Formation of Space-Charge Region

9 If a semiconductor diode has an initial uniform free electron concentration of \(5\times {10}^{21} m^{-3}\), and the depletion process completely removes these from a small volume of \(2\times {10}^{-15} m^{3}\), what will be the total number of free electrons depleted?

A \(2.5\times {10}^{6}\)

B \(1.0\times {10}^{7}\)

C \(1.0\times {10}^{6}\)

D \(2.5\times {10}^{7}\)

💡 Short Explanation

\(N=nV\)

✅ Correct Answer Explanation

\(N=(5\times {10}^{21})(2\times {10}^{-15})=10\times {10}^{6}=1.0\times {10}^{7}\)

❓ Why Other Options Are Incorrect

Obtained from incorrect multiplication.

🎯 Strategy

Used

Direct Calculation

Application:

Multiply concentration by volume.

Final Logic:

\(N=1.0\times {10}^{7}\)

🧠 Memory Trick

"Number = Concentration × Volume."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Depletion Region

10 Incorrect statement about the depletion region thickness

A It is generally of the order of one-tenth of a micrometre.

B Its thickness increases as the diffusion process continues before equilibrium.

C It becomes thick enough to completely halt the drift current.

D It encapsulates a space-charge region on either side of the metallurgical junction.

💡 Short Explanation

Drift current is not halted. Drift and diffusion currents balance at equilibrium.

✅ Correct Answer Explanation

The depletion region grows until the electric field becomes strong enough to produce a drift current equal and opposite to the diffusion current. Drift current continues to exist at equilibrium.

❓ Why Other Options Are Incorrect

A → Correct order of magnitude.
B → Correct behavior before equilibrium.
D → Correct definition.

🎯 Strategy

Used

Equilibrium Analysis

Application:

Understand depletion-layer growth.

Final Logic:

Drift current persists at equilibrium.

🧠 Memory Trick

"Equilibrium Means Balance, Not Zero Drift."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Depletion Layer and Equilibrium

11 The polarity of the barrier potential

A supports the further continuous flow of majority carriers across the junction

B opposes the further flow of carriers so that a condition of equilibrium exists

C is positive on the p-side and negative on the n-side

D depends entirely on the orientation of the external metallic contacts

💡 Short Explanation

Barrier potential develops naturally in a p-n junction. It opposes further diffusion of majority carriers. This establishes equilibrium.

✅ Correct Answer Explanation

The built-in electric field and barrier potential oppose the diffusion of majority carriers across the junction. Equilibrium is reached when drift current balances diffusion current.

❓ Why Other Options Are Incorrect

Option A → Barrier potential opposes, not supports, carrier flow.
Option C → Actual polarity is positive on n-side and negative on p-side.
Option D → Barrier potential is an internal property of the junction.

🎯 Strategy

Used

Conceptual Understanding

Application:

Identify the function of the barrier potential.

Final Logic:

Barrier potential prevents continuous majority-carrier diffusion.

🧠 Memory Trick

"Barrier Builds Balance."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Barrier Potential

12 The n-material in a p-n junction at equilibrium
1. has lost electrons
2. has acquired holes
3. is negative relative to the p-material
4. is positive relative to the p-material

A (A) and

B (B) and

C only

D only

💡 Short Explanation

Electrons diffuse from n-side to p-side. Positive donor ions remain behind. n-side becomes positive.

✅ Correct Answer Explanation

The n-region loses some electrons due to diffusion. The uncovered donor ions create a positive space charge, making the n-region positive relative to the p-region.

❓ Why Other Options Are Incorrect

Statement B → Holes do not accumulate in the n-region.
Statement C → n-region becomes positive, not negative.

🎯 Strategy

Used

Charge Distribution Analysis

Application:

Analyze charge formation near the junction.

Final Logic:

Loss of electrons leaves the n-region positively charged.

🧠 Memory Trick

"Lose Electrons → Become Positive."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Formation of Barrier Potential

13 When a small external voltage is applied under forward bias, the barrier potential is reduced ____, and only a ____ number of carriers possess enough energy to cross the junction.

A slightly, small

B significantly, large

C slightly, large

D significantly, small

💡 Short Explanation

Small forward bias slightly lowers the barrier. Only a limited number of carriers can cross initially.

✅ Correct Answer Explanation

For a small forward bias, the barrier potential decreases only slightly. Consequently, only a small fraction of majority carriers gain sufficient energy to cross the junction.

❓ Why Other Options Are Incorrect

B & C → Large carrier flow occurs only at higher forward voltages.
D → Contradictory statement.

🎯 Strategy

Used

Physical Interpretation

Application:

Relate small forward bias to carrier transport.

Final Logic:

Small voltage → small barrier reduction → small carrier crossing.

🧠 Memory Trick

"Small Bias, Small Crossing."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Forward Bias

14 Voltage drop statements in a biased diode
1. The applied voltage mostly drops across the depletion region.
2. The voltage drop across the p-side and n-side is negligible.
3. The depletion region has very low resistance.
4. The depletion region is highly resistive due to no free charges.

A (A), (B), (C)

B (B), (C), (D)

C (A), (B), (D)

D (A), (C), (D)

💡 Short Explanation

Depletion region contains almost no mobile carriers. Its resistance is very high. Most applied voltage appears across it.

✅ Correct Answer Explanation

Statements A, B, and D are correct. The depletion layer is highly resistive because it lacks free carriers. Therefore, almost the entire applied voltage appears across it.

❓ Why Other Options Are Incorrect

Statement C → Depletion region does not have low resistance.

🎯 Strategy

Used

Region Resistance Analysis

Application:

Determine where voltage drops in a diode.

Final Logic:

High resistance causes most voltage to drop across the depletion region.

🧠 Memory Trick

"No Carriers → High Resistance."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Effect of Biasing on Depletion Region

15 Match List I with List II regarding forward bias

List I	List II
I. p-side connection	P. Mostly across depletion region
II. n-side connection	Q. Decreases
III. Depletion width	R. Positive terminal
IV. Applied voltage drop	S. Negative terminal

A I-S, II-R, III-Q, IV-P

B I-R, II-S, III-Q, IV-P

C I-R, II-S, III-P, IV-Q

D I-S, II-R, III-P, IV-Q

💡 Short Explanation

p-side → positive terminal. n-side → negative terminal. Depletion width decreases. Voltage drops mainly across depletion layer.

✅ Correct Answer Explanation

Correct matching: I → R II → S III → Q IV → P

❓ Why Other Options Are Incorrect

Incorrect matching of battery connections and depletion width behavior.

🎯 Strategy

Used

Matching Concepts

Application:

Associate forward-bias characteristics.

Final Logic:

Only Option B correctly matches all items.

🧠 Memory Trick

"Forward Bias: P to Positive."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Forward Bias of p-n Junction

16 Correct statements about the barrier height under forward bias
1. The direction of the applied voltage opposes the built-in potential V₀.
2. The effective barrier height becomes (V₀ − V).
3. Increasing the applied voltage significantly increases the barrier height.
4. A reduced barrier height allows more carriers to have the required energy to cross.

A (A), (B), (D)

B (B), (C), (D)

C (A), (B), (C)

D (A), (C), (D)

💡 Short Explanation

Forward bias opposes the barrier. Barrier height decreases. Carrier flow increases.

✅ Correct Answer Explanation

Forward bias reduces the effective barrier height: \(V_{effective}=V_{0}-V\) As the barrier decreases, more carriers can cross the junction.

❓ Why Other Options Are Incorrect

Statement C → Forward bias reduces, not increases, barrier height.

🎯 Strategy

Used

Formula-Based Reasoning

Application:

Analyze forward-bias effects.

Final Logic:

A, B, and D are correct.

🧠 Memory Trick

"Forward Bias Lowers the Barrier."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Barrier Height Under Forward Bias

17 If the initial barrier height is V₀, and a reverse bias voltage V₁ and an additional series reverse bias V₂ are applied, the total effective barrier height will be

A (V₀ − V₁ − V₂)

B (V₀ + V₁ + V₂)

C −(V₀ + V₁ + V₂)

D (V₀ + V₁ − V₂)

💡 Short Explanation

Reverse bias adds to the barrier potential. Additional reverse voltages further increase it.

✅ Correct Answer Explanation

Under reverse bias: \(V_{effective}=V_{0}+V_{1}+V_{2}\) because all reverse-bias voltages reinforce the built-in barrier.

❓ Why Other Options Are Incorrect

Do not correctly represent reverse-bias addition.

🎯 Strategy

Used

Formula Recall

Application:

Apply reverse-bias barrier relation.

Final Logic:

Reverse voltages add to the built-in potential.

🧠 Memory Trick

"Reverse Bias = Add the Barriers."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Reverse Bias

18 If a diode has a barrier potential of \(V_{0}=0.6 V\), and experiences a reverse drift current of \(2.5 \mu A\), what will be the total charge passing through the junction in 4 seconds strictly due to this drift current?

A \(1.0\times {10}^{-5} C\)

B \(1.0\times {10}^{-4} C\)

C \(1.0\times {10}^{-6} C\)

D \(1.0\times {10}^{-7} C\)

💡 Short Explanation

Use: \(Q=It\)

✅ Correct Answer Explanation

\(I=2.5\times {10}^{-6} At=4 sQ=ItQ=(2.5\times {10}^{-6})(4)Q=1.0\times {10}^{-5} C\)

❓ Why Other Options Are Incorrect

Obtained from incorrect multiplication.

🎯 Strategy

Used

Direct Numerical Calculation

Application:

Use the current-charge relationship.

Final Logic:

\(Q=1.0\times {10}^{-5} C\)

🧠 Memory Trick

"Charge = Current × Time."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Reverse Saturation Current

19 Incorrect statement regarding minority carrier injection

A It takes place primarily under forward bias.

B Electrons from the n-side cross over and become minority carriers on the p-side.

C Holes from the p-side cross over and become minority carriers on the n-side.

D The minority carrier concentration at the junction edge becomes lower than far from the junction.

💡 Short Explanation

Forward bias injects carriers across the junction. Minority carrier concentration becomes highest near the junction.

✅ Correct Answer Explanation

Injected carriers accumulate near the depletion-region boundaries, causing the minority carrier concentration near the junction edge to increase significantly.

❓ Why Other Options Are Incorrect

A → Correct.
B → Correct electron injection process.
C → Correct hole injection process.

🎯 Strategy

Used

Carrier Injection Analysis

Application:

Understand carrier distribution near the junction.

Final Logic:

Minority carrier concentration increases near the junction.

🧠 Memory Trick

"Injection Raises Boundary Concentration."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Minority Carrier Injection

20 The total diode forward current

A is solely due to the drift of minority carriers

B is the sum of hole diffusion current and conventional current due to electron diffusion

C is purely a drift current of the order of a few µA

D is negligible due to the extremely high resistance of the p-side

💡 Short Explanation

Forward current mainly arises from diffusion. Both electrons and holes contribute.

✅ Correct Answer Explanation

Under forward bias, majority carriers cross the junction and diffuse into the opposite region. The total forward current is the sum of: Hole diffusion current Electron diffusion current

❓ Why Other Options Are Incorrect

A → Not solely drift current.
C → Forward current is predominantly diffusion current.
D → Forward-biased diode conducts significantly.

🎯 Strategy

Used

Current Component Analysis

Application:

Identify contributions to forward current.

Final Logic:

Forward current = electron diffusion current + hole diffusion current.

🧠 Memory Trick

"Forward Current = Electron + Hole Diffusion."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Current in a Forward-Biased p-n Junction

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