X-rays

Continuous X-rays (Bremsstrahlung):

• Mechanism: Electrons decelerate in the electric field of target atoms, converting kinetic energy to electromagnetic radiation
• Energy distribution:
  • Varies continuously from zero to maximum energy (eV)
  • Energy distribution is determined by:
    • Random collision parameters (impact parameter, energy loss per collision)
    • Electron's path through the target
  • Has a cutoff wavelength λ₂min = hc/eV

Characteristic X-rays:

• Mechanism: Two-step process:
  1. Incident electron ejects inner-shell electron (K, L, etc.)
  2. Outer electron transitions to fill vacancy, releasing excess energy as X-ray
• Energy distribution:
  • Discrete wavelengths corresponding to specific electronic transitions
  • Energies determined by:
    • Difference between atomic energy levels (EK-EL, EL-EM, etc.)
    • Atomic number of target material (follows Moseley's Law)
  • Appears as sharp peaks superimposed on continuous spectrum

Derivation of Moseley's Law from Bohr model:

1. For a transition from n=2 to n=1 in a Bohr atom with nuclear charge Ze: $\Delta E = Rhc(Z)^2\left(\frac{1}{1^2} - \frac{1}{2^2}\right) = \frac{3}{4}Rhc(Z)^2$

2. But inner electrons shield the nuclear charge, so effective charge is (Z-b)e: $\Delta E = \frac{3}{4}Rhc(Z-b)^2$

3. This energy equals the photon energy: $h\nu = \frac{3}{4}Rhc(Z-b)^2$

4. Solving for √ν: $\sqrt{\nu} = \sqrt{\frac{3Rc}{4}}(Z-b)$

5. Which gives Moseley's Law: $\sqrt{\nu} = a(Z-b)$ where $a = \sqrt{\frac{3Rc}{4}}$

Why b ≈ 1:

• For Kα X-rays, one electron remains in the K shell after ejection
• This K electron screens approximately one unit of nuclear charge
• The L electron making the transition "sees" an effective nuclear charge of approximately (Z-1)e
• Experimental measurements confirm b ≈ 1.37, slightly higher than theoretical prediction due to imperfect screening by the remaining K electron

This screening effect is essential for accurate prediction of characteristic X-ray frequencies.

Calculation for iron Kα X-ray:

Given:

• K shell binding energy = 7,100 eV
• L shell binding energy = 710 eV
• Z = 26 (iron)

1. The Kα X-ray energy equals the difference between K and L shell binding energies: Energy = EK - EL = 7,100 eV - 710 eV = 6,390 eV

2. Convert energy to wavelength using E = hc/λ: $\lambda = \frac{hc}{E} = \frac{1242 \text{ eV·nm}}{6390 \text{ eV}} = 0.194 \text{ nm}$

3. Check using Moseley's Law: If √ν = a(Z-b) where a ≈ 5.0×10⁷ (Hz)^(1/2) and b ≈ 1.37 √ν = 5.0×10⁷ × (26-1.37) = 5.0×10⁷ × 24.63 = 1.23×10⁹ Hz^(1/2) ν = 1.52×10¹⁸ Hz

   λ = c/ν = 3×10⁸ / 1.52×10¹⁸ = 0.197 nm (close to our calculated value)

This 0.194 nm X-ray is in the high-energy end of the spectrum and characteristic of iron.

X-ray diffraction angle calculation:

Given:

• Interplanar spacing d = 0.25 nm
• X-ray wavelength λ = 0.154 nm

Using Bragg's Law: 2d·sinθ = nλ, where n is an integer

For n = 1: 2(0.25 nm)·sinθ₁ = 1(0.154 nm) sinθ₁ = 0.154/(2×0.25) = 0.308 θ₁ = sin⁻¹(0.308) = 17.9°

For n = 2: 2(0.25 nm)·sinθ₂ = 2(0.154 nm) sinθ₂ = 2×0.154/(2×0.25) = 0.616 θ₂ = sin⁻¹(0.616) = 38.0°

For n = 3: 2(0.25 nm)·sinθ₃ = 3(0.154 nm) sinθ₃ = 3×0.154/(2×0.25) = 0.924 θ₃ = sin⁻¹(0.924) = 67.5°

For n ≥ 4: sinθ₄ would exceed 1, which is impossible.

Therefore, constructive interference would occur at angles of approximately 17.9°, 38.0°, and 67.5° corresponding to the first three orders of diffraction.

Soft vs. Hard X-rays:

Differentiation:

• Hard X-rays:

  • Shorter wavelengths (typically <0.1 nm)
  • Higher photon energies (typically >10 keV)
  • Greater penetrating power
  • Less easily absorbed by matter

• Soft X-rays:

  • Longer wavelengths (typically 0.1-10 nm)
  • Lower photon energies (typically 0.1-10 keV)
  • Less penetrating power
  • More easily absorbed by matter

Determining factors in a Coolidge tube:

1. Primary factor - Accelerating voltage (kVp):

   • Higher voltage → shorter cutoff wavelength → harder X-rays
   • Formula: λ₂min = hc/eV
   • Direct relationship between voltage and maximum photon energy

2. Secondary factors:

   • Target material (Z): Higher Z produces more characteristic X-rays, potentially making spectrum harder
   • Filtration: Thin metal filters (e.g., Al, Cu) can preferentially absorb soft X-rays
   • Tube design: Anode angle, window material affect beam quality

Note: Changing filament current affects X-ray intensity but not hardness/quality.

Cutoff wavelength mechanism:

The cutoff wavelength (λ₂min) represents the shortest possible wavelength in the X-ray spectrum produced at a given accelerating voltage, arising from:

1. A single electron converting ALL its kinetic energy (eV) into a single photon in one collision
2. This occurs when an electron transfers its entire kinetic energy to a photon without energy losses to the target

Quantum mechanical significance:

1. Demonstrates energy conservation at quantum level:

   • Energy transfer must occur in discrete amounts (quanta)
   • Maximum photon energy cannot exceed electron kinetic energy

2. Validates Einstein's photoelectric equation in reverse:

   • E = hν or E = hc/λ where E ≤ eV

3. Provides experimental evidence for:

   • Wave-particle duality of electromagnetic radiation
   • Quantization of energy transfer processes
   • Planck's constant (h) through the relationship λ₂min = hc/eV

4. Represents a fundamental physical limit independent of:

   • Target material
   • Tube design
   • Electron path through target

This quantum limit is precisely measurable and was historically important in establishing quantum mechanics.

Effects of operating parameter changes:

• Increasing filament voltage/current:

• Increases filament temperature
• Increases number of electrons emitted (thermionic emission)
• Results in higher X-ray intensity without changing X-ray energy spectrum
• Does not affect cutoff wavelength

• Increasing accelerating voltage (V):

• Increases electron kinetic energy (E = eV)
• Decreases minimum wavelength (λₘᵢₙ = hc/eV)
• Produces "harder" X-rays (more penetrating)
• Shifts the continuous spectrum toward higher energies

• Target material selection:

• Determines the wavelengths of characteristic X-rays
• Higher atomic number materials produce more efficient X-ray generation
• Affects heat dissipation requirements

Note: While intensity is controlled by filament current, the energy/penetrating power is controlled by accelerating voltage.

Characteristic X-rays are produced through the following process:

1. An energetic electron knocks out an electron from the K shell (n=1), creating a vacancy
2. An electron from a higher energy shell (L, M, or N) transitions to fill this vacancy
3. The energy difference between the two shells is released as an X-ray photon

Specific transitions produce specific X-ray lines:

• Kα: L→K transition (n=2→n=1)
• Kβ: M→K transition (n=3→n=1)
• Kγ: N→K transition (n=4→n=1)

Each element has characteristic energy differences between shells, producing unique X-ray wavelengths that can be used to identify elements (basis of X-ray spectroscopy).

X-ray diffraction in crystalline solids works through these steps:

1. X-rays strike a crystal at angle $\theta$ to parallel atomic planes
2. Each atom acts as a scattering center, re-radiating the X-rays
3. Scattered waves interfere with each other
4. Strong reflected beams occur only at specific angles where constructive interference happens

Specific angles produce strong reflections because:

• Constructive interference requires path difference = $n\lambda$
• This occurs only when $2d\sin\theta = n\lambda$ (Bragg's Law)
• At other angles, destructive interference cancels the waves
• The crystal's regular atomic spacing ($d$) determines these specific angles

This principle allows crystallographers to determine atomic arrangements by measuring diffraction patterns.

To find the angle for first-order diffraction:

1. Apply Bragg's Law: $2d\sin\theta = n\lambda$

2. Solve for $\theta$ when $n=1$:

   $\sin\theta = \frac{n\lambda}{2d} = \frac{1 \times 0.154\text{ nm}}{2 \times 0.254\text{ nm}} = \frac{0.154}{0.508} = 0.303$

3. Therefore: $\theta = \sin^{-1}(0.303) = 17.65°$

For first-order diffraction ($n=1$), X-rays will be strongly reflected at an angle of 17.65° from these atomic planes.

Note: If no strong reflection is observed at this angle, it suggests either the calculated plane spacing is incorrect or there might be systematic absences due to the crystal's structure factor.