Potential Well, Step & Barrier PYQs | IIT JAM Physics

Quantum Mechanics is one of the important and scoring sections of the IIT JAM Physics syllabus, and within it, Potential Well, Potential Step, and Potential Barrier form a core topic from which questions are asked almost every year. A strong conceptual understanding of these systems is essential, not only for solving numerical problems but also for interpreting the physical behavior of quantum particles.

Here, we focus on Previous Year Questions (PYQs) from IIT JAM, explained in a step-by-step and concept-oriented manner. Rather than memorizing formulas, focus on logical derivations and physical interpretation, which are crucial for tackling both direct and twisted questions in the JAM examination.

Introduction to Quantum Potential Systems

There are three fundamental one-dimensional potential configurations in quantum mechanics: the potential step, the potential well, and the potential barrier. These systems are foundational for understanding how quantum particles behave when encountering changes in potential energy, providing insights into phenomena like reflection, transmission, and tunneling.

The Infinite Potential Well (1D Box)

For a particle trapped in a 1D infinite potential well of width 'a' (from x=0 to x=a), where the potential V=0 inside and V=∞ outside, its behavior is described by specific wave functions and quantized energy levels.

• Wave Function (Ψ): The stationary state wave function for a particle inside the well is given by:
  Ψ(x) = √(2/a) * sin(nπx/a)
  Note: 'a' represents the width of the box, which can be denoted by any variable (e.g., L, 2L, etc.).

• Energy Eigenvalues (Eₙ): The allowed energy levels for the particle are quantized and depend on the quantum number 'n':
  Eₙ = n²π²ħ² / 2ma²
  Where 'm' is the mass of the particle.

• A crucial rule for the quantum number 'n' is that it must be a positive integer starting from 1. It can never be zero.
  n = 1, 2, 3, …

• n=1 corresponds to the Ground State.

• n=2 corresponds to the First Excited State.

• n=3 corresponds to the Second Excited State, and so on.

The Potential Step

The behavior of a particle encountering a potential step of height V₀ depends on whether its energy (E) is greater or less than V₀.

Here is a Comparative Analysis: Particle Energy vs. Step Height

The Potential Barrier

A potential barrier of height V₀ and finite width acts like a "speed breaker."

| Case | Description & Wave Functions | k₁ = √(2mE / ħ²) | 

The reflection coefficient (R) is close to 1 when E < V₀.

The probability of finding the particle within the classically forbidden region (where E < V) decreases exponentially with distance. This decay is governed by a characteristic length scale. 

| Electromagnetism | Skin Depth |

 The amplitude of an electromagnetic wave decays exponentially as it penetrates a conductor. The skin depth is the characteristic distance over which the amplitude falls to 1/e of its initial value.

| Electronics | Capacitor Discharging |

The charge on a discharging capacitor decreases exponentially with time. The time constant (τ) is related to the time it takes for the charge to fall to 1/e of its initial value.

| Nuclear Physics | Radioactivity |

The number of undecayed radioactive nuclei decreases exponentially over time. The half-life is the time required for the number of nuclei to reduce to half (1/2) of its initial amount.

Potential Well, Step & Barrier Solved Problems

To master Potential Well, Potential Step, and Potential Barrier, it is essential to apply theoretical concepts to actual problems. In this section, we solve carefully selected PYQs from various exams relevant to IIT JAM:

Problem 1: Energy Levels in Infinite Wells

Problem: The ground state energy of a particle in an infinite well of width L₁ equals the second excited state energy of a particle in another well of width L₂. Find the ratio L₁/L₂.

Solution: Eₙ = n²π²ħ² / 2mL². For ground state n=1. For the second excited state n=3.

(1)²π²ħ² / 2m(L₁)² = (3)²π²ħ² / 2m(L₂)²

1 / (L₁)² = 9 / (L₂)²

(L₂ / L₁)² = 9 => L₂ / L₁ = 3. 

Therefore, L₁ / L₂ = 1/3.

Problem 2: Comparing Electron and Muon in a Well

Problem: An electron and a muon are trapped in identical 1D potential wells. Given a muon is much heavier than an electron, compare their energies and wave functions.

Solution: Energy levels are inversely proportional to mass: E ∝ 1/m. 

Since m_μ ≈ 210 * m_e, then E_electron > E_muon for any given quantum state 'n'. 

Wave functions have the same mathematical form √(2/a)sin(nπx/a), but the energies differ due to mass.

Problem 3: Mixed States and Expectation Values (Harmonic Oscillator)

Problem: A harmonic oscillator is in a mixed state |ψ⟩ = (1/√3)|ψ₀⟩ - (1/√6)|ψ₁⟩ + (1/√2)|ψ₂⟩. Analyze energy measurements.

Solution:

1. Probabilities: P(E₀) = |1/√3|² = 1/3. P(E₁) = |-1/√6|² = 1/6. P(E₂) = |1/√2|² = 1/2.

2. Possible Energies: Only E₀ = 1/2 ħω, E₁ = 3/2 ħω, E₂ = 5/2 ħω can be measured. A measurement yielding 5/3 ħω is false.

3. Expectation Value ⟨E⟩:
   ⟨E⟩ = (1/3)(1/2 ħω) + (1/6)(3/2 ħω) + (1/2)(5/2 ħω) = (1/6 + 3/12 + 5/4) ħω = (2+3+15)/12 ħω = 20/12 ħω = 5/3 ħω.

Problem 4: Degeneracy in a 3D Cubic Well

Problem: Find the degeneracy of the fifth excited state of a particle in a 3D cubic potential well.

Solution: E ∝ (nx² + ny² + nz²).

• Ground: (1,1,1) -> sum=3 (Degeneracy=1)

• 1st Excited: (1,1,2) -> sum=6 (Degeneracy=3)

• 2nd Excited: (1,2,2) -> sum=9 (Degeneracy=3)

• 3rd Excited: (1,1,3) -> sum=11 (Degeneracy=3)

• 4th Excited: (2,2,2) -> sum=12 (Degeneracy=1)

• 5th Excited: (1,2,3) -> sum=14. Since nx, ny, nz are all different, degeneracy is 3! = **6**.

Problem 5: Uncertainty Principle in a 1D Well

Problem: For a particle in the ground state of a 1D well from -L/2 to L/2, determine the uncertainty in momentum (Δp).

Solution: Δp = √(<p²> - <p>²). For a stationary state in a 1D well, ⟨p⟩ = 0.

So, Δp = √<p²>. Also, E = <p²>/2m, so ⟨p²⟩ = 2mE.

For ground state (n=1) in well of width L (from -L/2 to L/2), E₁ = π²ħ² / 2mL².

Δp = √(2m * π²ħ² / 2mL²) = √(π²ħ² / L²) = **πħ / L**.

Problem 6: Superposed State in an Infinite Square Well

Problem: A particle in an infinite square well of length L is in a superposed state of ψ₁ (P=1/3) and ψ₂ (P=2/3). Identify correct statements.

Solution:

1. ⟨p⟩ = 0: True for any stationary state or superposition in a 1D infinite well.

2. Δx = 0: False. Position uncertainty cannot be zero for a confined particle.

3. ⟨E⟩ = 3h² / 8mL²: True. Eₙ = n²π²ħ² / 2mL².
   ⟨E⟩ = (1/3)E₁ + (2/3)E₂ = (1/3)(1²π²ħ²/2mL²) + (2/3)(2²π²ħ²/2mL²) = (1/3 + 8/3)(π²ħ²/2mL²) = 3(π²ħ²/2mL²).
   Using ħ = h/2π, ⟨E⟩ = 3(h²/4π²)(π²/2mL²) = 3h² / 8mL².

4. Δp = (√3 * h) / 2L: True. Δp = √<p²>. Since ⟨E⟩ = <p²>/2m, ⟨p²⟩ = 2m⟨E⟩ = 2m(3h²/8mL²) = 3h²/4L².
   Δp = √(3h²/4L²) = (√3 * h) / 2L.

Problem 7: Reflection Coefficient at a Step Potential

Problem: A quantum particle of mass m and energy E encounters a step potential V₀. Analyze the reflection coefficient R.

Solution:

• If E > V₀: Classically, R=0. Quantum mechanically, there's partial reflection and transmission (R is between 0 and 1).

• If E < V₀: Classically, R=1. Quantum mechanically, there's a very high probability of reflection, so R ≈ 1. There is an exponentially decaying penetration into the barrier, but significant transmission (like R=0.5) is not possible.

Problem 8: Fermions in a 1D Potential Well

Problem: Five non-interacting electrons in a 1D well of width a at 0 Kelvin. Find the energy of the highest occupied state.

Solution: Electrons are fermions, obeying the Pauli Exclusion Principle (max 2 electrons per state). At 0K, states are filled from lowest energy.

• n=1: 2 electrons

• n=2: 2 electrons

• n=3: The 5th electron occupies this state.
  Highest occupied state is n=3, with energy E₃ = 9π²ħ² / 2ma².

Problem 9: Energy Levels of a 3D Cubic Box

Problem: For a particle in a cubic box of side L, ground state energy E₀. Find the energy of the first excited state in terms of E₀.

Solution: E = (nx² + ny² + nz²) * (π²ħ² / 2mL²).

• Ground State (E₀): (1,1,1) => E₀ = (1² + 1² + 1²) * (π²ħ² / 2mL²) = **3 * (π²ħ² / 2mL²)**.

• First Excited State: (1,1,2) (or its permutations) => E_first_excited = (1² + 1² + 2²) * (π²ħ² / 2mL²) = **6 * (π²ħ² / 2mL²)**.

• Therefore, E_first_excited = 2 * [3 * (π²ħ² / 2mL²)] = **2E₀**.

Problem 10: Probability in a 1D Infinite Well

Problem: Given ground state ψ(x) = √(2/L) sin(πx/L) for a particle in a 1D box (0 to L), find the probability of finding the particle between x=0 and x=L/2.

Solution:

1. Symmetry Method: The probability density |ψ(x)|² for the ground state is symmetric around x=L/2. Since total probability is 1, the probability in the first half (0 to L/2) is 1/2.

2. Integration Method: P = ∫[0 to L/2] (2/L) sin²(πx/L) dx. Using sin²θ = (1-cos2θ)/2, the integral evaluates to 1/2.

Problem 11: Neutron Penetration into a Potential Barrier

Problem: Neutrons (E = 8 MeV) incident on a potential step (V₀ = 48 MeV). Calculate distance x where probability density decreases by a factor of 100 from its value at x=0.

Given: ħc = 200 MeV-fm, mc² = 1 GeV = 1000 MeV.

Solution:

1. Probability Density: P(x) = |ψ(x)|² = |C|² * e^(-2k₂x). Condition: P(x) = P(0)/100 => e^(-2k₂x) = 10⁻².

2. Solving for x: -2k₂x = ln(10⁻²) = -2 ln(10) => x = ln(10) / k₂ ≈ 2.303 / k₂.

3. Calculate k₂: k₂² = 2m(V₀ - E) / ħ² = [2(mc²)(V₀ - E)] / (ħc)².
   V₀ - E = 48 - 8 = 40 MeV.
   k₂² = [2 * (1000 MeV) * (40 MeV)] / (200 MeV-fm)² = 80000 MeV² / 40000 (MeV-fm)² = 2 fm⁻².
   k₂ = √2 fm⁻¹ ≈ 1.414 fm⁻¹.

4. Final x: x = 2.303 / 1.414 ≈ **1.63 fm**.

General Advice & Problem-Solving Strategy

Problems based on Potential Well, Potential Step, and Potential Barrier often test both conceptual understanding and mathematical accuracy. A systematic approach is therefore essential to avoid common errors and save time during the examination:

• Strategic Calculation: Develop the presence of mind to use given values (like mc² and ħc) efficiently to simplify complex calculations.

• Memorize Standard Results: Memorizing key expectation values (e.g., <x> = middle of the box, <p> = 0 for stationary states) saves critical exam time.

• Classically Forbidden Regions: When a particle's energy (E) is less than the potential (V), it's in a classically forbidden region. The reflection coefficient is nearly 1, and the probability of finding the particle within the barrier decreases exponentially.