Ionising and thermal radiation are different exposures

Thermal and ionising describe different aspects

Thermal radiation describes emission associated with temperature. Ionising radiation has sufficient energy in its interactions to remove electrons from relevant atoms or molecules. These descriptions are not mutually exclusive classifications: a spectrum and its interactions determine what damage can occur.

A coating can warm under infrared light while an oxide in a semiconductor device accumulates trapped charge under an energetic-particle exposure. A thermometer alone cannot tell these histories apart. Begin with three questions: where did the radiation come from, what energy did its photons or particles carry, and what interactions occurred in this particular material?

Thermal describes an emission process associated with a body’s temperature. Ionising describes an interaction capable of removing a bound electron from an atom or molecule. Thus the words are not opposite ends of a single spectrum. Much radiation from everyday-temperature objects has photons too weak to ionise ordinary atoms in a single interaction, but the high-energy part of a sufficiently hot thermal spectrum can be ionising. Conversely, non-ionising absorption can still heat, excite carriers or initiate some chemical changes.

One photon’s energy is different from beam intensity

For a photon in vacuum, E = hf = hc/λ. Here h is Planck’s constant, f frequency, c the speed of light and λ wavelength. With E in electronvolts and λ in micrometres, E ≈ 1.23984/λ. One electronvolt is 1.602176634 × 10⁻¹⁹ J. Here beam intensity means irradiance, energy arriving per area per time; increasing the number of photons at a fixed wavelength does not increase each photon’s energy.

To make an energy comparison precise, invent an isolated target with a single-photon ionisation threshold E₀ = 10 eV, initially in its ground state. A 10 μm photon carries 0.124 eV, a 0.50 μm photon 2.48 eV and a 0.10 μm photon 12.4 eV. Only the last has enough energy for this chosen single-photon process. Above-threshold energy makes the event energetically possible; absorption probability and competing interactions still matter. This is not a claim that every incident photon ionises the target.

The threshold is a stated property of this invented target, not a universal boundary at 124 nm. Removing an electron from an isolated atom, ejecting one across a metal’s surface work function and promoting an electron between bands in a solid are different energy questions. A semiconductor can acquire additional mobile carriers without ejecting an electron into vacuum. Strong-field multiphoton processes are outside the single-photon comparison.

A thermal spectrum has a tail, not a single photon energy

An ideal blackbody at temperature T emits a continuous Planck spectrum. The familiar peak wavelength is not a cutoff: some emitted energy lies at shorter wavelengths and therefore higher photon energies. Real surfaces have wavelength-dependent emissivity, and windows, filters and geometry can change the spectrum arriving at a specimen.

For the comparison below, divide the emitted spectral energy by the total emitted energy. Let g_T(E) dE be the fraction in the photon-energy interval E to E + dE. In consistent energy units, g_T(E) = 15E³/{π⁴(k_B T)⁴[exp(E/(k_B T)) − 1]}, where k_B is Boltzmann’s constant. Its integral from zero to infinity is one. If E and k_B T are in eV, g_T has units eV⁻¹. This is a distribution of energy, not a distribution of photon counts.

This per-energy curve follows from the wavelength Planck law by changing variables λ = hc/E and including |dλ/dE| = hc/E². Simply relabelling the wavelength axis would give a wrong curve. For a chosen threshold E₀, put x₀ = E₀/(k_B T). The emitted energy fraction above it is F = (15/π⁴) ∫ from x₀ to infinity of x³/(exp x − 1) dx.

For E₀ = 10 eV, the ideal 6000 K emitter has F ≈ 5.20 × 10⁻⁶, whereas the ideal 30000 K emitter has F ≈ 0.428. The first is very small, not exactly zero. Neither number is an absorbed fraction or an ionisation yield. The target’s absorption and cross sections remain needed. Both plotted spectra have unit total area; equal-area ideal blackbodies at these temperatures emit different total powers, in the ratio (30000/6000)⁴ = 625.

Deposited energy can first become charge or chemical change

Energetic photons can transfer energy through photoelectric absorption or Compton scattering, producing energetic electrons. Those electrons can excite and ionise along their paths. An incident charged particle can likewise deposit energy through electronic interactions. Photon energy or particle kinetic energy alone does not specify where the energy is deposited or how densely interactions are clustered. Atomic displacement is a separate structural pathway developed in the next lesson.

A material-specific example is a silicon-based device with an insulating oxide. Ionising energy creates electron–hole pairs in the oxide. Some pairs recombine; an electric field can separate the remaining charges. Their different transport and trapping behaviour can leave oxide charge and interface states that change the voltage needed to turn a transistor on. Temperature, bias and processing history influence these processes. The same temperature or energy per mass therefore does not specify the same surviving trapped charge.

This is a mechanism example, not a universal radiation-response law for every oxide or device. Likewise, ultraviolet exposure can produce chain scission or crosslinking in polymers; calling an exposure non-ionising does not establish chemical harmlessness. In space-material evidence, vacuum, reactive atomic oxygen and specimen temperature can accompany radiation. A changed coating must be attributed using the actual combined exposure, rather than radiation’s name alone.

Heat is not the only response

Ionising exposure can create charge or chemical changes without being described adequately by specimen temperature alone. Conversely, absorbed radiation can heat a material. Specify particles or photons, energy distribution, dose measure, rate and environment as relevant. Absorbed dose is deposited energy per mass; its SI unit, the gray (Gy), equals 1 J/kg. The absorbing material and spatial region must be identified. Equal dose does not guarantee equal surviving defects or equal property loss.

Here absorbed dose in gray refers to ionising radiation. For ordinary infrared heating, report the absorbed radiant energy and heat balance rather than assigning a dosimetric label just because J/kg is dimensionally available. Locally, absorbed dose is D = dε̄/dm, the mean energy imparted per mass of a specified medium. For a finite region, its mass-averaged dose is ε̄_region/m_region. A mean is not necessarily the dose at every point.

Incident energy per area, often called energy fluence, is different from energy actually imparted per mass. A beam attenuated out of a narrow forward detector may have scattered rather than deposited all the missing energy locally. Fluorescent photons and secondary electrons can also carry energy across the region’s boundary. Attenuation and local absorption therefore cannot generally be interchanged.

Dose rate is dose per time, for example Gy/s. Even when total dose is equal, different delivery histories can allow different amounts of charge transport, chemical reaction or recovery while exposure continues. A property comparison should identify the absorbing medium and region, radiation type and energy distribution, rate, temperature, bias where relevant, environment and elapsed observation time.

A stated deposition model exposes the spatial average

Consider an original one-dimensional model of a homogeneous fictional medium M. A normally incident monoenergetic ionising-photon field brings total mean energy U₀ through area A. Let κ be a constant pure-absorption coefficient and ρ the density. Assume no reflection, scattering or fluorescent escape; all absorbed energy is imparted within the same thin layer, with negligible secondary-particle range and lateral losses. These deliberately strong assumptions make attenuation equal to local deposition. They are not a general photon-transport approximation near real interfaces.

The remaining beam energy at depth z is U(z) = U₀ exp(−κz). A layer dz absorbs κU(z) dz and has mass ρA dz. Division gives D(z) = κU₀ exp(−κz)/(ρA). The numerator has units J/m and the denominator kg/m, so D is J/kg. Integrating over thickness L gives absorbed energy U_abs = U₀[1 − exp(−κL)] and mass-averaged dose D̄ = U_abs/(ρAL).

For ρ = 2000 kg/m³, A = 1.00 cm², L = 2.00 mm, κ = 1000 m⁻¹ and U₀ = 0.0800 J, the model gives D̄ ≈ 173 Gy. The front-face limit is 400 Gy and the back-face limit 54.1 Gy. A uniform-density slab makes the thickness average equal to the mass average here. Reporting only 173 Gy would hide the strong gradient. Real scattering, secondary transport, composition and geometry can require a different depth-dose calculation.

Temperature follows an energy balance, not dose alone

If all deposited energy becomes heat, no energy is stored in chemical change or persistent defects, heat capacity c_p is constant and no phase change occurs, an adiabatic specimen has ΔT = U_abs/(mc_p) = D̄/c_p after internal thermal equilibration. For the slab with c_p = 500 J/(kg·K), this is only 0.346 K. A small temperature rise does not bound trapped charge or chemical change. Conversely, energy stored in nonthermal changes and heat exchange invalidate that simple calorimetric inference.

To see the role of delivery time, assume instead a spatially uniform specimen temperature with total heat capacity C = mc_p, initially at its surroundings’ temperature T₀. A constant effective thermal conductance G removes heat at rate Gθ, where θ = T − T₀. With constant deposited heating power P_heat, energy conservation gives C dθ/dt = P_heat − Gθ. Solving with θ(0) = 0 gives θ(t) = (P_heat/G)[1 − exp(−t/τ)], with τ = C/G. This lumped model requires fast internal thermal equilibration compared with the heating and cooling times, constant properties and the stated linear heat-loss approximation.

After deposition stops, θ decays exponentially towards zero in this model. The accumulated absorbed dose is still a description of the exposure history; the return to T₀ does not by itself prove that trapped charge, altered bonds or structural defects have recovered. Their relaxation has its own mechanisms and time scales.

Test an attribution using quantities that discriminate between mechanisms. A thermal explanation needs temperature history and a heat balance; an ionisation explanation needs radiation and medium-specific energy deposition plus a relevant charge, spectroscopic or electrical response. A matched heating comparison and controlled environment can narrow alternatives, but a similar temperature trace alone cannot establish identical microscopic histories. State which property was measured and when: resistance, optical transmission and mechanical strength are different outcomes.

Worked examples

Try the reasoning

1. Is every thermal photon ionising? Explain why a spectrum’s peak does not settle this question.

   Show the answer and reasoning

   No. A thermal spectrum contains a range of photon energies.

   For the specified target, only photons meeting its relevant energy condition can cause the chosen single-photon ionisation. A peak is a maximum of a spectral density, not an end of the distribution; the high-energy tail can be very small without being zero.

2. Does an energy dose alone identify every damage process? Two specimens each receive 100 Gy in the same identified medium but under different bias and dose-rate histories. What conclusion is justified?

   Show the answer and reasoning

   Their mass-averaged imparted energies per mass match; equal surviving defects or equal property loss is not established.

   100 Gy means 100 J/kg in the specified region. Charge separation, recombination, reaction and recovery depend on history and state. A spatial average can also hide different depth distributions.

3. In the single-photon model, a new isolated target has threshold 6.0 eV. Find the boundary wavelength. Can doubling the intensity of 250 nm light make each photon meet that threshold?

   Show the answer and reasoning

   The boundary is about 207 nm. A 250 nm photon has about 4.96 eV, so doubling its intensity does not satisfy the chosen one-photon requirement.

   λ₀ = 1239.841984/6 nm = 206.640 nm. E(250 nm) = 1239.841984/250 = 4.95937 eV. Increasing photon number at fixed wavelength changes incident energy rate, not photon energy. Multiphoton processes are excluded by the problem.

4. A filtered field arriving at a specimen contains 0.050 J in total. Its measured spectrum puts 20% of that energy above a chosen threshold. Suppose the material absorbs exactly half of that above-threshold energy. Find that absorbed amount. Is it enough to calculate an ionisation count?

   Show the answer and reasoning

   0.0050 J is absorbed from the above-threshold part; the ionisation count is undetermined.

   Multiply 0.050 × 0.20 × 0.50. Energy can be distributed among different photon energies, secondary excitations and other channels. A yield or interaction model is still needed; dividing by a threshold energy without such a model would invent a count.

5. Double only the ideal slab’s thickness from 2.00 mm to 4.00 mm, retaining κ, ρ, A and total U₀. Predict, then calculate, the change in total imparted energy and mass-averaged dose. Does the front-face model limit change?

   Show the answer and reasoning

   Imparted energy rises to about 0.0785 J, but the mean falls to about 98.2 Gy. The front limit remains 400 Gy.

   κL becomes 4, U_abs = 0.0800(1 − exp(−4)) = 0.0785347 J and mass doubles to 0.000800 kg. Their ratio is 98.1684 Gy. Extra thickness captures more remaining beam energy but adds relatively weakly irradiated mass. D(0) does not contain L. These conclusions use the ideal local-deposition model.

6. A narrow forward detector measures only 30% of the incident photon energy after a real slab. May one assign the missing 70% to local absorbed energy and divide by slab mass?

   Show the answer and reasoning

   No; forward attenuation alone does not close the local energy balance.

   Scattered and fluorescent photons can leave in other directions, and secondary particles can cross the boundary. The pure-absorption assumptions would need justification or replacement by appropriate transport and energy-deposition evidence.

7. A uniform specimen receives 200 Gy and has c_p = 800 J/(kg·K). Under the adiabatic all-heat, constant-property assumptions, find its equilibrated rise. What can be inferred if its measured rise is smaller?

   Show the answer and reasoning

   The conditional rise is 0.250 K. A smaller observed rise alone does not identify the missing energy pathway.

   ΔT = D/c_p = 200/800. Heat loss, nonthermal energy storage, incomplete spatial equilibration or measurement limitations could matter. Temperature alone does not separate these alternatives, so it cannot validate or refute the dose without the rest of the model.

What to carry forward

• Thermal origin and ionising capability are different descriptors; specify the spectrum and material interaction.
• A per-energy thermal distribution, a photon-count distribution and an absorbed-energy distribution are different quantities.
• Absorbed dose belongs to an identified medium and region. Its mass average can conceal a strong gradient.
• Temperature depends on deposition, heat capacity and heat exchange; persistent charge, chemistry and structure require their own evidence.