Module V: Quantum Physics - Question Solutions

Question 1: de Broglie Hypothesis of Matter Waves

Question

Explain the de Broglie hypothesis of matter waves.

de Broglie Hypothesis (1924)

Statement:

Louis de Broglie proposed that just as light exhibits both wave and particle nature (wave-particle duality), all matter (particles with mass) also exhibits wave-like properties. Every moving particle has an associated wave called a matter wave or de Broglie wave.

de Broglie Wavelength

The wavelength associated with a moving particle is given by:

λ = h/p = h/(mv)

de Broglie wavelength equation

where:

λ = de Broglie wavelength (m)
h = Planck's constant = 6.626 × 10⁻³⁴ J·s
p = momentum of the particle (kg·m/s)
m = mass of the particle (kg)
v = velocity of the particle (m/s)

Derivation of de Broglie Equation

From Einstein's and Planck's Relations

Step 1: For a photon, Einstein's mass-energy relation gives:

E = mc²

Step 2: Planck's quantum relation:

E = hν = hc/λ

Step 3: Equating the two expressions:

mc² = hc/λ

Step 4: Solving for λ:

λ = h/(mc) = h/p

Step 5: de Broglie extended this to all particles with momentum p = mv:

λ = h/(mv)

de Broglie Wavelength for an Electron

For an electron accelerated through potential V:

Kinetic Energy = eV = ½mv²

Therefore: v = √(2eV/m)

Substituting in de Broglie equation:

λ = h/√(2meV)

Substituting values (h = 6.626×10⁻³⁴ J·s, m = 9.1×10⁻³¹ kg, e = 1.6×10⁻¹⁹ C):

de Broglie Wavelength for Electron

λ = 12.27/√V Å

where V is in volts and λ is in angstroms

Significance of de Broglie Hypothesis

Established wave-particle duality for all matter
Explained electron orbits in Bohr's model (standing wave condition)
Foundation for quantum mechanics and Schrödinger equation
Experimentally verified by Davisson-Germer experiment (electron diffraction)
Led to development of electron microscopes

Question 2: Properties of Matter Waves

Question

State the properties of matter waves.

Properties of Matter Waves (de Broglie Waves):

1. Wavelength inversely proportional to momentum: λ = h/p. Heavier or faster particles have shorter wavelengths.
2. Not electromagnetic waves: Matter waves are probability waves associated with particles, not electromagnetic radiation.
3. Wave-particle duality: Matter exhibits both wave and particle properties depending on the experimental setup.
4. Phase velocity greater than c: The phase velocity vₚ = ω/k can exceed the speed of light, but this doesn't violate relativity as it doesn't carry information.
5. Group velocity equals particle velocity: The group velocity vg = dω/dk equals the particle's actual velocity.
6. Cannot be observed for macroscopic objects: Due to extremely small wavelength (λ ∝ 1/m), wave nature is negligible for large masses.
7. Diffraction and interference: Matter waves exhibit diffraction and interference patterns (verified experimentally).
8. Wavelength independent of charge: The de Broglie wavelength depends only on momentum, not on the particle's charge.
9. Associated with probability: The intensity of matter wave (|ψ|²) represents the probability density of finding the particle.
10. Require quantum mechanical description: Matter waves are described by the wave function ψ governed by Schrödinger equation.

vₚ × vg = c²

Relationship between phase velocity and group velocity

Important Relationships

Phase velocity: vₚ = ω/k = E/p = c²/v (greater than c)
Group velocity: vg = dω/dk = v (equals particle velocity)
For non-relativistic particles: vg = 2vₚ

Question 3: Schrödinger's Time Independent Wave Equation

Question

Derive Schrödinger's Time Independent wave equation for matter waves.

Derivation

Step-by-Step Derivation

Step 1: Consider a particle moving along x-direction. The classical wave equation is:

∂²ψ/∂x² = (1/v²)(∂²ψ/∂t²)

Step 2: Assume the wave function has the form:

ψ(x,t) = ψ₀ e^(i(kx - ωt))

Step 3: For a time-independent (stationary) solution:

ψ(x,t) = ψ(x) e^(-iωt)

Step 4: Calculate the second derivative with respect to x:

∂²ψ/∂x² = d²ψ/dx² × e^(-iωt)

Step 5: From de Broglie relation, p = h/λ = ℏk, where k = 2π/λ

k² = p²/ℏ²

Step 6: The total energy of a particle:

E = KE + PE = p²/(2m) + V

Step 7: Solving for p²:

p² = 2m(E - V)

Step 8: Substituting into k²:

k² = 2m(E - V)/ℏ²

Step 9: The wave equation in terms of k gives:

d²ψ/dx² = -k²ψ

Step 10: Substituting the expression for k²:

d²ψ/dx² = -[2m(E - V)/ℏ²]ψ

Schrödinger's Time Independent Wave Equation (1D)

d²ψ/dx² + (2m/ℏ²)(E - V)ψ = 0

Or equivalently:

-(ℏ²/2m)(d²ψ/dx²) + Vψ = Eψ

Three-Dimensional Form

∇²ψ + (2m/ℏ²)(E - V)ψ = 0

where ∇² = ∂²/∂x² + ∂²/∂y² + ∂²/∂z² is the Laplacian operator.

Significance

This equation gives the allowed energy levels (eigenvalues) of a quantum system
Solutions ψ are called eigenfunctions or stationary states
Used for bound state problems (atoms, molecules, quantum wells)
E represents total energy, V is potential energy

Question 4: Particle in 1D Infinite Potential Well

Question

An electron is bound in an 1D potential well of width 2 Å but of infinite height. Find its energy values in the ground state and in first excited state. (Given: Planck constant is 6.634×10⁻³⁴ J-Sec)

Energy Formula for Particle in Infinite Well

Infinite Potential Well

For a particle in an infinite potential well of width L, the allowed energy levels are:

Eₙ = n²h²/(8mL²)

where n = 1, 2, 3, ... (quantum number)

Given Data:

Width of well, L = 2 Å = 2 × 10⁻¹⁰ m
Planck's constant, h = 6.634 × 10⁻³⁴ J·s
Mass of electron, m = 9.1 × 10⁻³¹ kg

Ground State Energy (n = 1):

E₁ = (1)² × h²/(8mL²)

E₁ = (6.634 × 10⁻³⁴)²/(8 × 9.1 × 10⁻³¹ × (2 × 10⁻¹⁰)²)

E₁ = (44.01 × 10⁻⁶⁸)/(8 × 9.1 × 10⁻³¹ × 4 × 10⁻²⁰)

E₁ = (44.01 × 10⁻⁶⁸)/(291.2 × 10⁻⁵¹)

E₁ = 0.1511 × 10⁻¹⁷ J

E₁ = 1.511 × 10⁻¹⁸ J

Converting to electron volts (1 eV = 1.6 × 10⁻¹⁹ J):

E₁ = (1.511 × 10⁻¹⁸)/(1.6 × 10⁻¹⁹) eV

Ground State Energy E₁ = 9.44 eV ≈ 9.4 eV

First Excited State Energy (n = 2):

E₂ = (2)² × E₁ = 4 × E₁

E₂ = 4 × 1.511 × 10⁻¹⁸ J

E₂ = 6.044 × 10⁻¹⁸ J

In electron volts:

E₂ = (6.044 × 10⁻¹⁸)/(1.6 × 10⁻¹⁹) eV

First Excited State Energy E₂ = 37.78 eV ≈ 37.8 eV

Key Observations

Energy levels are quantized: Eₙ ∝ n²
Ground state (n=1) has non-zero energy (zero-point energy)
E₂ = 4E₁, E₃ = 9E₁, etc.
Energy gap increases with n

Question 5: Electron Cannot Reside Inside Nucleus

Question

Prove that electron cannot reside inside the nucleus of an atom.

Proof Using Heisenberg's Uncertainty Principle

Heisenberg's Uncertainty Principle:

Δx × Δp ≥ ℏ/2 ≈ h/(4π)

The product of uncertainty in position and momentum of a particle cannot be less than ℏ/2.

Step 1: Consider the nucleus size

The radius of a typical nucleus is approximately:

r ≈ 10⁻¹⁴ m to 10⁻¹⁵ m

Let's take Δx ≈ 10⁻¹⁴ m (diameter ~ 2 × 10⁻¹⁴ m)

Step 2: Calculate minimum uncertainty in momentum

From uncertainty principle:

Δp ≥ ℏ/(2Δx) = h/(4πΔx)

Δp ≥ (6.626 × 10⁻³⁴)/(4 × 3.14 × 10⁻¹⁴)

Δp ≥ (6.626 × 10⁻³⁴)/(12.56 × 10⁻¹⁴)

Δp ≥ 5.27 × 10⁻²¹ kg·m/s

Step 3: Calculate minimum momentum

The minimum momentum of the electron must be at least equal to the uncertainty:

p_min ≈ Δp ≈ 5.27 × 10⁻²¹ kg·m/s

Step 4: Calculate minimum kinetic energy

Using relativistic or non-relativistic formula. First, let's check if relativistic treatment is needed by comparing p with m₀c:

m₀c = 9.1 × 10⁻³¹ × 3 × 10⁸ = 2.73 × 10⁻²² kg·m/s

Since p >> m₀c, we need relativistic treatment. Using:

E = pc (ultra-relativistic approximation)

E = 5.27 × 10⁻²¹ × 3 × 10⁸

E = 1.58 × 10⁻¹² J

Step 5: Convert to electron volts

E = (1.58 × 10⁻¹²)/(1.6 × 10⁻¹⁹) eV

E ≈ 9.9 × 10⁶ eV ≈ 10 MeV

Conclusion

If an electron were confined inside the nucleus (Δx ~ 10⁻¹⁴ m), it would need a minimum kinetic energy of approximately 10 MeV.

However, experimental observations show that:

Beta particles (electrons) emitted from nuclei have energies of only 2-3 MeV maximum
The binding energy of electrons in atoms is only a few eV to keV

This enormous discrepancy proves that electrons cannot exist inside the nucleus.

Additional Evidence

No electron energy levels exist within the nucleus
Beta decay produces new electrons rather than releasing existing ones
The electron's Compton wavelength (~10⁻¹² m) is much larger than nuclear dimensions

Question 6: Electron and Photon Momentum and Energy

Question

An electron and a photon each have a wavelength of 2Å. What are their momentum and energies?

Given Data:

Wavelength, λ = 2 Å = 2 × 10⁻¹⁰ m
Planck's constant, h = 6.626 × 10⁻³⁴ J·s
Speed of light, c = 3 × 10⁸ m/s
Mass of electron, mₑ = 9.1 × 10⁻³¹ kg

Part A: Momentum (Same for Both)

Momentum Calculation:

From de Broglie relation, momentum is related to wavelength by:

p = h/λ

p = (6.626 × 10⁻³⁴)/(2 × 10⁻¹⁰)

p = 3.313 × 10⁻²⁴ kg·m/s

Momentum (for both electron and photon): p = 3.313 × 10⁻²⁴ kg·m/s

Part B: Energy of Photon

Photon Energy:

E_photon = hc/λ = pc

E_photon = (6.626 × 10⁻³⁴ × 3 × 10⁸)/(2 × 10⁻¹⁰)

E_photon = (19.878 × 10⁻²⁶)/(2 × 10⁻¹⁰)

E_photon = 9.939 × 10⁻¹⁶ J

In electron volts:

E_photon = (9.939 × 10⁻¹⁶)/(1.6 × 10⁻¹⁹) eV = 6212 eV ≈ 6.21 keV

Photon Energy: E = 9.939 × 10⁻¹⁶ J = 6.21 keV

Part C: Energy of Electron

Electron Energy (Kinetic):

For a non-relativistic electron:

E_electron = p²/(2m)

E_electron = (3.313 × 10⁻²⁴)²/(2 × 9.1 × 10⁻³¹)

E_electron = (10.976 × 10⁻⁴⁸)/(18.2 × 10⁻³¹)

E_electron = 6.03 × 10⁻¹⁸ J

In electron volts:

E_electron = (6.03 × 10⁻¹⁸)/(1.6 × 10⁻¹⁹) eV = 37.7 eV

Electron Energy: E = 6.03 × 10⁻¹⁸ J = 37.7 eV

Comparison

Property	Photon	Electron
Wavelength	2 Å	2 Å
Momentum	3.313 × 10⁻²⁴ kg·m/s	3.313 × 10⁻²⁴ kg·m/s
Energy	6.21 keV	37.7 eV

The photon has much higher energy than the electron for the same wavelength because E = pc for photon but E = p²/2m for electron.

Question 7: de Broglie Wavelength Calculation

Question

Calculate the wavelength of de-Broglie waves associated with mass 1 kg moving with a speed of 10³ m/sec.

Given Data:

Mass, m = 1 kg
Velocity, v = 10³ m/s = 1000 m/s
Planck's constant, h = 6.626 × 10⁻³⁴ J·s

Formula:

λ = h/(mv)

Calculation:

λ = (6.626 × 10⁻³⁴)/(1 × 10³)

λ = 6.626 × 10⁻³⁷ m

de Broglie wavelength λ = 6.626 × 10⁻³⁷ m

Observation

This wavelength (≈ 10⁻³⁷ m) is extremely small - much smaller than even the size of a proton (~10⁻¹⁵ m). This is why wave properties of macroscopic objects are not observable in everyday life. The wave nature becomes significant only for particles with very small mass (like electrons).

Question 8: Heisenberg's Uncertainty Principle - Position Accuracy

Question

An electron has a speed of 400 m/sec with uncertainty of 0.01%. Find the accuracy in its position.

Given Data:

Speed of electron, v = 400 m/s
Uncertainty in speed = 0.01% of v
Mass of electron, m = 9.1 × 10⁻³¹ kg
Planck's constant, h = 6.626 × 10⁻³⁴ J·s

Step 1: Calculate uncertainty in velocity

Δv = 0.01% × v = (0.01/100) × 400

Δv = 0.04 m/s

Step 2: Calculate uncertainty in momentum

Δp = m × Δv

Δp = 9.1 × 10⁻³¹ × 0.04

Δp = 3.64 × 10⁻³² kg·m/s

Step 3: Apply Heisenberg's Uncertainty Principle

Δx × Δp ≥ h/(4π)

Δx ≥ h/(4π × Δp)

Δx ≥ (6.626 × 10⁻³⁴)/(4 × 3.14 × 3.64 × 10⁻³²)

Δx ≥ (6.626 × 10⁻³⁴)/(45.72 × 10⁻³²)

Δx ≥ (6.626 × 10⁻³⁴)/(4.572 × 10⁻³¹)

Δx ≥ 1.45 × 10⁻³ m

Minimum uncertainty in position Δx ≥ 1.45 × 10⁻³ m = 1.45 mm ≈ 0.145 cm

Interpretation

Even with a very small uncertainty (0.01%) in velocity, the position of the electron cannot be determined more accurately than about 1.45 mm. This demonstrates the fundamental quantum mechanical limitation on simultaneous measurement of position and momentum.