JEE Physics Atoms Notes Set 02

Production of X-rays

X-rays are electromagnetic waves with wavelengths from about 0.01 Å to 100 Å, between ultraviolet light and γ-rays. They were found by Röntgen, who noticed that a fast electron beam striking matter gives out a highly penetrating radiation. They travel at the speed of light, carry no charge and are not deflected by electric or magnetic fields.

In the Coolidge tube a tungsten filament, heated by a low-voltage current, releases electrons by thermionic emission. A high voltage V (tens of kilovolts) between the cathode and the metal target (anode) accelerates the electrons, which hit the target with kinetic energy eV. The tube is evacuated to a very low pressure so that the electrons are not slowed by gas atoms, and the target has a high melting point because most of the electron energy (over 99%) turns into heat. The target is often cooled by circulating water or oil.

Two controls decide the properties of the beam. The filament current fixes how many electrons are released each second, so it controls the intensity. The accelerating voltage fixes the energy of the electrons, so it controls the quality (hardness) of the X-rays. Hard X-rays have short wavelengths and high penetration and soft X-rays have longer wavelengths and are absorbed more.

Continuous Spectrum and Cut-off Wavelength

When electrons slow down in the field of the target nuclei they radiate photons of all energies up to the electron energy, which gives a continuous spectrum. The most energetic photon appears when an electron gives up all its energy in a single event, so eV = hc/λ_min, and therefore \( \lambda_{min} = \frac{hc}{eV} = \frac{12400}{V}\ \text{Å} \), with V in volts. Also ν_max = eV/h.

The cut-off wavelength depends only on the voltage and not on the target material or the filament current. If V is doubled, λ_min is halved.

Example. At 25 kV, λ_min = 12400/25000 = 0.496 Å. At 50 kV the maximum photon energy is 50 keV and the maximum frequency is 50000 × 1.6 × 10⁻¹⁹/6.63 × 10⁻³⁴ = 1.2 × 10¹⁹ Hz.

Characteristic X-rays and Moseley's Law

If the bombarding electron has enough energy it can knock an electron out of an inner shell (K, L, M ...) of a target atom. An electron from a higher shell then falls into the vacancy and a photon with the energy difference is emitted. This radiation has sharp lines whose wavelengths are fixed by the target element, so the lines are called characteristic X-rays.

A jump from the L shell (n = 2) to the K shell (n = 1) gives the K_α line, from the M shell (n = 3) to the K shell gives K_β, and from M to L gives L_α. The K_β photon has more energy than the K_α photon, so its wavelength is shorter. The characteristic lines appear only when the tube voltage exceeds the energy needed to ionise that shell, and their wavelengths do not change with the voltage.

Moseley found that the square root of the frequency of a given characteristic line is a straight-line function of the atomic number: \( \sqrt{\nu} = a(Z - b) \). For the K_α line b = 1, and the law can be written \( \frac{1}{\lambda} = \frac{3R}{4}(Z-1)^2 \). Hence ν is proportional to (Z − 1)² and λ is proportional to 1/(Z − 1)². Moseley's work showed that the atomic number, not the atomic mass, is the correct basis of the periodic table.

Example. If K_α of molybdenum (Z = 42) has λ = 0.71 Å, then for copper (Z = 29) λ = 0.71 × (41/28)² = 1.52 Å.

Diffraction of X-rays

The spacing between atomic planes in a crystal is a few angstroms, of the same order as the X-ray wavelength, so a crystal acts as a three-dimensional diffraction grating. Strong reflection occurs when the path difference between rays from neighbouring planes is a whole number of wavelengths, which is Bragg's law: 2d sin θ = nλ, where d is the plane spacing, θ is the glancing angle between the beam and the plane, and n = 1, 2, 3, ... is the order.

Since sin θ cannot exceed 1, diffraction is possible only if λ ≤ 2d, so the longest wavelength that a crystal can diffract is 2d. This technique is used to find crystal structures and the spacings between atoms.

Example. For d = 3 × 10⁻¹⁰ m and a first-order reflection at θ = 30°, λ = 2 × 3 × 10⁻¹⁰ × 0.5 = 3 × 10⁻¹⁰ m.

Absorption of X-rays

When X-rays pass through matter, the loss of intensity in a thin layer is proportional to the intensity and to the thickness: dI = −μI dx. Integrating gives I = I₀e^−μx, where μ is the absorption coefficient. The half-value thickness, at which the intensity falls to one half, is x_½ = ln 2/μ = 0.693/μ.

The absorption coefficient is larger for denser materials and materials of higher atomic number, and it is smaller for harder (shorter wavelength) X-rays. This is why bone shows up on an X-ray picture while flesh does not, and why lead is used for shielding.

Example. If the intensity falls to one fourth after 3.5 mm of a metal, then μ = ln 4/3.5 mm = 0.40 mm⁻¹.

Properties and Uses

X-rays affect photographic plates, make some substances fluoresce, ionise gases and release photoelectrons from metals. They show diffraction but are not refracted by ordinary lenses to any useful extent. They are used for medical imaging and cancer therapy, for detecting cracks in castings and welds, for checking luggage at airports and for studying crystal structure and the structure of molecules such as DNA. Over-exposure damages living tissue.