Rigid vs Non-Rigid Body

A rigid body is an ideal model of a solid in which the distance between any two particles remains constant, even when external forces are applied. It ignores deformation and is used to simplify analysis of translational and rotational motion.

A non-rigid (deformable) body is one in which the distance between particles can change under applied forces. Such bodies can stretch, compress, or permanently deform depending on the material properties.

Derivation of Rotational Kinetic Energy

Consider a rigid body rotating with angular velocity ω about a fixed axis. Let the body consist of particles of mass mᵢ at distances rᵢ from the axis.

Linear speed:

vᵢ = rᵢ ω

Kinetic energy of a particle:

KEᵢ = ½ mᵢ vᵢ²
KEᵢ = ½ mᵢ (rᵢ ω)²
KEᵢ = ½ mᵢ rᵢ² ω²

Total rotational kinetic energy:

KE_rot = Σ ½ mᵢ rᵢ² ω²
KE_rot = ½ ω² Σ mᵢ rᵢ²

Since:

I = Σ mᵢ rᵢ²

Therefore:

KE_rot = ½ I ω²

Condition for Conservation of Energy

Mechanical energy is conserved when only conservative forces act on a system.

Conditions:

• No non-conservative forces (friction, air resistance, etc.)
• System is isolated (no external work input or loss)

Calculation of Rotational Velocity

Given:

R = 0.4 m
I = 1.2 kg·m²
m = 2 kg
h = 1.5 m
g = 9.8 m/s²

Energy conservation:

mgh = ½ mv² + ½ I ω²

v = Rω

Substitute:

mgh = ½ m(Rω)² + ½ I ω²
mgh = ½ ω² (mR² + I)

So:

ω² = (2mgh) / (mR² + I)

Substitution:

mR² = 2 × (0.4)² = 0.32 kg·m²
mR² + I = 0.32 + 1.2 = 1.52 kg·m²

2mgh = 2 × 2 × 9.8 × 1.5 = 58.8 J

ω² = 58.8 / 1.52
ω ≈ 6.22 rad/s

The classical free electron model, proposed by Drude and refined by Lorentz, describes the electrical and thermal properties of metals based on the following premises:

• Free Electron Gas: Valence electrons are assumed to detach from their parent atoms and move freely throughout the metal, forming a "gas" of electrons.
• Fixed Ion Cores: The positive ion cores are considered stationary, forming a uniform positive background potential within the metal lattice.
• Negligible Electron-Electron Interactions: Interactions between individual electrons are neglected.
• Elastic Collisions: Electrons undergo instantaneous, elastic collisions only with the fixed ion cores, which scatter them and change their direction of motion. The time between collisions is known as the relaxation time tau.
• Classical Statistics: The motion of electrons is governed by classical mechanics, and their energy distribution follows Maxwell-Boltzmann statistics.

This model successfully explained phenomena such as Ohm's Law, the Wiedemann-Franz Law (ratio of thermal to electrical conductivity), and the general concept of electrical and thermal conduction in metals. However, its major shortcomings included the incorrect prediction of electronic specific heat capacity and the temperature dependence of resistivity, which were later addressed by quantum mechanical models.

(a) Oxidation
Oxidation is the process of growing a thin layer of silicon dioxide (SiO₂) on the surface of a silicon wafer.

• Process: Silicon wafers are heated to high temperatures (typically 900-1200°C) in an atmosphere containing oxygen (dry oxidation) or water vapor (wet oxidation).
  • Dry Oxidation: Si + O₂ → SiO₂
  • Wet Oxidation: Si + 2H₂O → SiO₂ + 2H₂
• Purpose:
  • Acts as an excellent electrical insulator, isolating different components.
  • Serves as a mask against dopants during diffusion or ion implantation.
  • Passivates the silicon surface, reducing leakage currents.

(b) Doping
Doping is the intentional introduction of impurities (dopants) into an intrinsic semiconductor (e.g., silicon) to modify its electrical conductivity.

• Process:
  • Ion Implantation: Dopant atoms are ionized and accelerated by an electric field into the silicon wafer, controlling their depth and concentration.
  • Diffusion: Wafers are heated in a furnace containing a gaseous source of dopants. Dopant atoms diffuse into the silicon crystal lattice due to thermal energy.
• Purpose: To create p-type semiconductors (by adding trivalent impurities like Boron, creating holes) or n-type semiconductors (by adding pentavalent impurities like Phosphorus or Arsenic, creating free electrons), forming junctions essential for transistors and diodes.

Photolithography
Photolithography is a critical patterning process used to transfer geometric patterns from a photomask to the surface of a semiconductor wafer.

• Steps:
  1. Photoresist Application: A light-sensitive polymer (photoresist) is uniformly spun onto the wafer surface.
  2. Exposure: The wafer, covered with photoresist, is exposed to UV light (or X-rays/electron beams) through a photomask. The mask contains the desired pattern.
     • Positive Photoresist: Exposed areas become soluble.
     • Negative Photoresist: Exposed areas become insoluble.
  3. Development: A chemical developer dissolves either the exposed or unexposed photoresist areas, leaving a patterned resist layer.
  4. Etching: The unprotected areas of the underlying material (e.g., SiO₂) are removed by chemical or plasma etching, transferring the pattern into the wafer material.
  5. Photoresist Removal: The remaining photoresist is stripped off, leaving the etched pattern.
• Purpose: To define the regions where subsequent processes (like doping, oxidation, or material deposition) will occur, enabling the creation of complex integrated circuits.

• Amplitude (A)
  The distance between the endpoints of the motion is twice the amplitude.
  Distance = 2A
  50 cm = 2A
  A = 50 cm / 2 = 25 cm
  A = 0.25 m

• Frequency (f)
  The time taken to move between the endpoints (from one extreme to the other) is half a period (T/2).
  T/2 = 0.2 s
  Period (T) = 0.2 s * 2 = 0.4 s
  Frequency (f) = 1/T
  f = 1 / 0.4 s
  f = 2.5 Hz

• Force Constant (k)
  For a mass-spring system, the period (T) is given by the formula:
  T = 2π√(m/k)
  Rearranging for k:
  T² = 4π²(m/k)
  k = (4π²m) / T²
  Given mass (m) = 250 g = 0.250 kg
  k = (4 * π² * 0.250 kg) / (0.4 s)²
  k = (9.8696 * 0.250) / 0.16
  k = 9.8696 / 0.16
  k ≈ 61.69 N/m

• Given:

  • Potential difference, V = 100 V
  • Charge of alpha particle, q = 3.2 × 10^-19 C
  • Mass of proton, m_p = 6.70-x 10^-19 kg (Typographical error in the question. Assuming the question intended the mass of an alpha particle, m_α, to be approximately 6.70 × 10^-27 kg, which is close to 4 times the mass of a proton (4 * 1.67 × 10^-27 kg). We will proceed with m_α = 6.70 × 10^-27 kg.)

• Principle:
  The work done by the electric field on the alpha particle as it moves through the potential difference is converted into its kinetic energy.
  Work done (W) = Change in kinetic energy (ΔKE)

• Formulas:
  Work done, W = qV
  Kinetic energy, KE = ½mv²

• Calculation:
  Since the alpha particle is released from rest (initial velocity u = 0), its initial kinetic energy is zero.
  The work done by the electric field is equal to the final kinetic energy gained by the particle:
  qV = ½m_αv²

  Rearranging to solve for final velocity, v:
  v² = (2qV) / m_α
  v = √((2qV) / m_α)

  Substitute the given values:
  v = √((2 × 3.2 × 10^-19 C × 100 V) / (6.70 × 10^-27 kg))
  v = √((6.4 × 10^-17 J) / (6.70 × 10^-27 kg))
  v = √(0.95522 × 10^10 m²/s²)
  v = √(9.5522 × 10^9 m²/s²)
  v ≈ 9.77 × 10^4 m/s

• Final Velocity:
  The velocity of the alpha particle when it reaches the other plate is approximately 9.77 × 10^4 m/s.

Heisenberg’s Uncertainty Principle is used to determine the minimum uncertainty in momentum.

Formula:

Δx · Δp ≥ h / (4π)

Given:

Δx = 1 Å = 1 × 10⁻¹⁰ m
h = 6.626 × 10⁻³⁴ J·s

Calculation:

Δp = h / (4πΔx)

Δp = (6.626 × 10⁻³⁴) / (4π × 1 × 10⁻¹⁰)

Δp = (6.626 × 10⁻³⁴) / (12.566 × 10⁻¹⁰)

Δp = 0.527 × 10⁻²⁴ kg·m/s

Δp ≈ 5.27 × 10⁻²⁵ kg·m/s

Result:

Minimum uncertainty in momentum = 5.27 × 10⁻²⁵ kg·m/s

Hall Voltage Calculation

• Given Data:

  • Current (I) = 50 A
  • Thickness (t) = 0.5 cm = 0.005 m
  • Width (w) = 2 cm = 0.02 m
  • Magnetic field (B) = 1.5 T
  • Free electron concentration (n) = 8.4 × 10^28 electrons/m^3
  • Elementary charge (e) = 1.6 × 10^-19 C

• Formula for Hall Voltage:
  The Hall voltage (V_H) across the width of the slab is given by:
  V_H = (I * B) / (n * e * t)

• Calculation:
  V_H = (50 A * 1.5 T) / (8.4 × 10^28 electrons/m^3 * 1.6 × 10^-19 C * 0.005 m)
  V_H = 75 / (8.4 * 1.6 * 0.005 * 10^(28-19))
  V_H = 75 / (0.0672 * 10^9)
  V_H = 75 / (6.72 * 10^7)
  V_H = 1.116 × 10^-6 V

• Magnitude of Hall Voltage:
  The magnitude of the Hall voltage across the width of the slab is 1.116 µV (or 1.116 × 10^-6 V).

The length of the unit cube of the structure can be determined using the formula relating density, atomic weight, number of atoms per unit cell, and Avogadro's number.

1. Formula for Density (ρ):
   ρ = (n * M) / (V * N_A)
   Where:

   • n = number of atoms per unit cell
   • M = atomic weight
   • V = volume of the unit cell
   • N_A = Avogadro's number (6.022 × 10²³ atoms/mole)

2. Known Values:

   • Structure: Body-Centered Cubic (BCC), thus n = 2 atoms/unit cell
   • Atomic weight (M) = 23 g/mole
   • Density (ρ) = 0.971 g/cm³
   • Avogadro's number (N_A) = 6.022 × 10²³ atoms/mole
   • For a cubic unit cell, Volume (V) = a³, where 'a' is the length of the unit cube.

3. Rearranging the formula to solve for 'a':
   a³ = (n * M) / (ρ * N_A)
   a = [(n * M) / (ρ * N_A)]^(1/3)

4. Calculation:
   a³ = (2 atoms/unit cell * 23 g/mole) / (0.971 g/cm³ * 6.022 × 10²³ atoms/mole)
   a³ = 46 / (5.847362 × 10²³) cm³/unit cell
   a³ = 7.8679 × 10⁻²³ cm³/unit cell
   a = (7.8679 × 10⁻²³) ^ (1/3) cm
   a = 4.286 × 10⁻⁸ cm

The length of the unit cube (lattice parameter) for sodium is 4.286 × 10⁻⁸ cm.