Radiation Damage, Swelling, DPA and Activation Calculations in Nuclear Materials

These are working notes on a set of calculations that appear again and again when dealing with irradiated materials. The topic looks intimidating at first because it mixes material science, nuclear physics and a rather large number of units. The underlying calculations however are mostly bookkeeping: count atoms, count reactions, keep track of time, convert units carefully and only then interpret what the result means physically.

The focus here is to provide a short intuitive summary on four connected questions:

• How do energetic particles damage a crystal lattice?
• How do we calculate displacements per atom (dpa)?
• How do gases and voids cause swelling?
• How do activation, decay chains and transmutation calculations work?

For real world applicatoin one usually needs simulation tools like SRIM, QCalc, FISPACT-II, ORIGEN), OpenMC and similar programs because real irradiation problems contain many isotopes, many reaction channels and energy-dependent nuclear data. The goal of this summary is to understand the simple calculations well enough that these tools do not look like black boxes.

• The Physical Picture
• Displacement Threshold Energy
• Energy Transfer in a Collision
• What dpa means
• Calculating dpa from Neutron Flux
• Dpa from SRIM Ion Irradiation
• From Point Defects to Material Property Changes
• Swelling in structural materials
• Helium and Grain Boundary Embrittlement
• Fuel Swelling is Different from Void Swelling
• Bubble Pressure and the Laplace Term
• Worked Swelling Example
• Radioactive Decay
• Activation
• U-238 to Pu-239 Conversion
• Decay Chains and Bateman Equations
• What the Common Tools are for
• Calculation Checklist
• References

The Physical Picture

Most engineering materials are made from atoms arranged in a solid. In many metals and ceramics the atoms are not placed randomly but sit on a crystal lattice. One can imagine a regular array of preferred atomic positions. At room temperature the atoms vibrate around these positions, but they do not normally leave them. Irradiation changes this. A neutron, ion, proton, alpha particle or fission fragment can transfer kinetic energy to one of the atoms in the solid. If the transferred energy is small, the target atom only vibrates more strongly and the energy is dissipated as heat. If the transferred energy is large enough, the atom is knocked out of its lattice position.

The missing atom site is called a vacancy. The displaced atom, if it ends up squeezed between normal lattice positions, is called an interstitial. Together they form a Frenkel pair:

[ \mathrm{Frenkel\ pair} = \mathrm{vacancy} + \mathrm{interstitial} ]

This is the smallest unit of displacement damage. Note that displacement damage is not the same thing as activation. A material can be displaced without becoming radioactive, and it can become activated without suffering much displacement damage. In a reactor both can happen at the same time, so it is useful to separate the ideas:

• Displacement damage changes the arrangement of atoms in the solid.
• Activation changes nuclei into radioactive nuclei.
• Transmutation changes one element or isotope into another.
• Radioactive decay changes unstable nuclei over time.

These notes go through each of these with the same basic method: first take a note what is being counted, then write down the rate, then integrate over time.

Displacement Threshold Energy

To permanently displace an atom from its lattice site, the transferred recoil energy has to exceed a material-dependent threshold. This is called the displacement threshold energy $E_d$

For many metals, a useful order-of-magnitude value is

[ E_d \approx 25\text{--}50 \mathrm{eV} ]

This number is much larger than the thermal energy of atoms at room temperature. At room temperature,

[ k_B T \approx 0.026\,\mathrm{eV} ]

so normal thermal vibration is far too weak to knock atoms permanently out of place. Irradiation damage is therefore a non-equilibrium process. It creates defect concentrations that can be far above the normal thermal-equilibrium vacancy concentration.

The first atom knocked out by an incoming particle is called the primary knock-on atom (PKA). If this PKA has enough energy, it knocks out further atoms. Those atoms can knock out more atoms. The result is a displacement cascade.

Energy Transfer in a Collision

For a simple elastic collision, the maximum energy that a projectile of mass $M$ and kinetic energy $E$ can transfer to a stationary target atom of mass $m$ is

[ T_{\max} = \frac{4Mm}{(M+m)^2} E ]

This already explains an important qualitative fact. Energy transfer is most efficient when the masses are similar. A very light particle hitting a very heavy atom cannot transfer all of its energy in one collision. A particle with a mass similar to the target atom can transfer a much larger fraction. This is also the reason water is used as moderator in many nuclear facilities - and also the reason water is an excellent shiedling material.

As a simple example, consider a $1\mathrm{MeV}$ neutron hitting an iron atom. A neutron has mass number about 1 while iron has mass number about 56. We use $M=1$, $m=56$ and $E=1\mathrm{MeV}$:

[ T_{\max} = \frac{4\cdot 1\cdot 56}{(1+56)^2} \cdot 1 \mathrm{MeV} \approx 0.069 \mathrm{MeV} ]

Thus the maximum recoil energy is about $69\mathrm{keV}$. For isotropic scattering a useful average estimate is approximately half the maximum recoil energy, about $35\mathrm{keV}$. Since a displacement may require only about $25\mathrm{eV}$, this one recoil atom can produce many further displacements. A very simple cascade estimate is

[ \nu \approx \frac{T}{2E_d} ]

where $\nu$ is the number of displaced atoms produced by a recoil of energy $T$. With $T=35 \mathrm{keV}$ and $E_d=25 \mathrm{eV}$:

[ \nu \approx \frac{35000}{2\cdot 25} = 700 ]

This is a simplified Kinchin-Pease style estimate. It is not a perfect model of a real cascade. In real materials, some energy is lost to electronic excitation, some defects recombine immediately, and the displacement threshold depends on direction in the crystal. Still, it is a very useful first estimate.

What dpa means

The standard unit for material displacement exposure is dpa, displacements per atom:

[ \mathrm{dpa} = \frac{\text{number of displacement events produced}}{\text{number of atoms in the material}} ]

A dose of $1\mathrm{dpa}$ means that, on average, one displacement event has been produced per atom. It does not mean that every atom is currently displaced. This distinction matters a lot. Most vacancies and interstitials recombine:

[ V + I \rightarrow \text{perfect lattice} ]

Other defects migrate to sinks such as surfaces, grain boundaries and dislocations. Only a fraction of the produced damage survives as retained microstructural damage. Therefore dpa should be read as a damage production measure.

Calculating dpa from Neutron Flux

For neutron irradiation one often uses a displacement cross section. A cross section is an effective area that represents the probability for a nuclear process. Microscopic cross sections are commonly measured in barns:

[ 1 \mathrm{barn}=10^{-24} \mathrm{cm^2} ]

If the neutron flux is $\phi$ and the displacement cross section is $\sigma_d$, then the $\mathrm{dpa}$ rate is

[ \partial_t d = \phi \sigma_d ]

The units show why this works:

[ \begin{aligned} \lbrack \phi \rbrack &= \frac{1}{\mathrm{cm^2 s}} \\ \lbrack \sigma_d \rbrack &= \mathrm{cm^2} \end{aligned} ]

so

[ \lbrack\phi\sigma_d\rbrack = \frac{1}{\mathrm{s}} ]

The accumulated dose after time $t$ is

[ d = \phi \sigma_d t ]

As an example, take

[ \begin{aligned} \phi &= 5\times10^{13} \mathrm{n cm^{-2} s^{-1}} \\ \sigma_d &= 1.25\times10^{-21} \mathrm{cm^2} \\ t &= 5 \mathrm{yr} \approx 1.6\times10^8 \mathrm{s} \end{aligned} ]

Then

[ d = (5\times10^{13})(1.25\times10^{-21})(1.6\times10^8) \approx 10 \mathrm{dpa} ]

This calculation is very compact because all the complicated energy dependence has been hidden inside $\sigma_d$. In a real reactor the neutron spectrum matters, so one generally integrates over neutron energy:

[ \partial_t d = \int \phi(E) \sigma_d(E) \mathrm{d}E ]

For a first hand calculation, however, using a spectrum-averaged displacement cross section is often enough.

Dpa from SRIM Ion Irradiation

Neutron irradiation is difficult to model, slow and often produces radioactive specimens. For material research one often uses ion irradiation to produce displacement damage more quickly. SRIM, the Stopping and Range of Ions in Matter, is a common tool for estimating the ion range and vacancy production profile.

A typical SRIM damage output may look like

[ D_\mathrm{SRIM} = 0.085 \frac{\mathrm{vacancies}}{\mathrm{ion}\cdot\mathrm{A}} ]

This means that at a particular depth, each incoming ion produces 0.085 vacancies per Angstrom of path length. To convert this into dpa, compare two quantities in the same thin slice:

• vacancies produced per $\mathrm{cm^2}$,
• atoms available per $\mathrm{cm^2}$.

Suppose the irradiation uses singly charged ions with $I=100\mathrm{nA}$ over $A=1\mathrm{cm^2}$ for $t=2\mathrm{h}=7200\mathrm{s}$ and the target atom density is $N=8.5\times10^{22} \mathrm{atoms cm^{-3}}$. The current counts charge per time. For singly charged ions,

[ I = e \partial_t N_i ]

so

[ \begin{aligned} \partial_t N_i &= \frac{I}{e} \\ &= \frac{100\times10^{-9}}{1.602\times10^{-19}} \\ &= 6.24\times10^{11} \mathrm{ions s^{-1}} \end{aligned} ]

The fluence is the number of ions per area:

[ \begin{aligned} \Phi &= \frac{\partial_t N_i t}{A} \\ &= \frac{(6.24\times10^{11})(7200)}{1} \\ &= 4.49\times10^{15} \mathrm{ions\,cm^{-2}} \end{aligned} ]

Now we need the number of target atoms in a 1 Angstrom thick layer under one square centimeter. Since $1 mathrm{A}=10^{-8} \mathrm{cm}$ we get

[ \begin{aligned} N_\mathrm{slice} &= N\cdot10^{-8} \\ &= 8.5\times10^{14} \mathrm{atoms\,cm^{-2}\,A^{-1}} \end{aligned} ]

The vacancies produced per square centimeter in that slice are

[ \begin{aligned} D_\mathrm{SRIM}\Phi &= (0.085)(4.49\times10^{15}) \\ &= 3.82\times10^{14} \mathrm{vacancies\,cm^{-2}\,A^{-1}} \end{aligned} ]

Thus the peak dpa is

[ \begin{aligned} d_\mathrm{peak} &= \frac{D_\mathrm{SRIM}\Phi}{N\cdot10^{-8}} \\ &= \frac{3.82\times10^{14}}{8.5\times10^{14}} \\ &\approx 0.45 \mathrm{dpa} \end{aligned} ]

The most common mistake in this calculation is forgetting that SRIM used Angstroms while the atom density used cubic centimeters.

Ion dpa and neutron dpa should also not be interpreted too casually. Ion irradiation produces a strongly depth-dependent damage profile. Neutron damage is usually much more uniform through small specimens. Ion irradiation can also introduce implanted ions and very high dose rates.

Ion irradiation is a powerful simulation method, but it is not automatically identical to neutron irradiation.

From Point Defects to Material Property Changes

Once vacancies and interstitials are produced, several things can happen:

• vacancies and interstitials recombine
• defects migrate to grain boundaries, dislocations or free surfaces
• interstitials cluster into dislocation loops
• vacancies cluster into voids
• gas atoms stabilize bubbles
• solutes diffuse and form radiation-induced segregation or precipitates

A beginner friendly way to interpret irradiation hardening is to think about dislocation motion. Metals deform plastically when dislocations move. Irradiation creates obstacles to that motion. Defect clusters, loops and precipitates pin or slow dislocations. Therefore the yield strength often increases:

[ \Delta \sigma_y > 0 ]

Higher yield strength is not automatically better. Irradiated metals often lose ductility. In body-centered cubic steels, irradiation can shift the ductile-to-brittle transition temperature upward. This is one of the central concerns for reactor pressure vessel steels - the vessel must remain tough under operating and accident conditions.

Temperature is important because defects have to move before they can form larger structures. Very cold materials may retain many small defects. At intermediate temperatures, vacancies become mobile enough to form voids. At higher temperatures, thermal creep and helium embrittlement can become important.

Swelling in structural materials

Swelling means that the material volume increases:

[ \frac{\Delta V}{V} > 0 ]

In structural metals under irradiation, swelling is often caused by vacancy clustering. If vacancies collect into three-dimensional clusters, they form cavities or voids. The void volume adds to the macroscopic volume of the material.

This is especially important because even a few percent swelling can matter in engineering components. A component does not have to melt or crack visibly to fail its function. It may simply change dimension, lose clearances, generate internal stresses or distort an assembly. There is also a difference between uniform and differential swelling. If an entire object swells uniformly, the dimensional change may be manageable. If one region swells more than another, internal stresses and bending can develop. Differential swelling is often more damaging than the average swelling value suggests.

Fast reactors are particularly sensitive to swelling because fast neutrons create energetic displacement cascades and high accumulated dpa. Austenitic stainless steels can show substantial void swelling under suitable temperature and dose conditions. Ferritic and ferritic/martensitic steels generally swell less, partly because their body-centered cubic structure and defect transport behavior promote more efficient recombination and sink absorption.

Helium and Grain Boundary Embrittlement

Helium is a special problem. It can be produced by nuclear reactions such as $(n,\alpha)$, where a neutron reaction emits an alpha particle, which is a helium nucleus. Helium can also be implanted directly during ion irradiation.

Helium is nearly insoluble in most metals. It prefers to collect at defects, voids and grain boundaries. Once helium atoms collect, they stabilize bubbles. At grain boundaries these bubbles can reduce cohesion between grains and promote intergranular fracture.

This is especially important in fusion and spallation environments, where helium production per dpa can be much higher than in many fission reactor environments. A material may therefore perform well under one neutron spectrum but poorly under another, even at the same nominal dpa.

Fuel Swelling is Different from Void Swelling

Nuclear fuel swelling is related to irradiation damage, but the mechanism is not identical to void swelling in structural steel. In $\mathrm{UO}_2$ fuel, fission splits heavy nuclei into fission products. Some fission products are solids, others are noble gases, mainly xenon and krypton. Xenon and krypton have low solubility in $\mathrm{UO_2}$. They tend to form gas atoms, bubbles and eventually interconnected porosity. This affects the fuel in several ways:

• the fuel volume increases
• the fuel-cladding gap can close
• the rod internal pressure can increase
• gas release can carry radioactivity if the cladding fails
• thermal conductivity decreases, which increases fuel temperature

A first fission gas calculation starts with the number of fissions and the fission yield. Assuming $N_f = 10^{21}$ fissions occur and the cumulative xenon yield is $Y_\mathrm{Xe}=6.5\%=0.065$, then the number of xenon atoms is

[ N_\mathrm{Xe} = Y_\mathrm{Xe}N_f = 0.065\cdot10^{21} = 6.5\times10^{19} ]

To use the ideal gas law, convert atoms to moles:

[ \begin{aligned} n_\mathrm{Xe} &= \frac{N_\mathrm{Xe}}{N_A} \\ &= \frac{6.5\times10^{19}}{6.022\times10^{23}} \\ &= 1.08\times10^{-4} \mathrm{mol} \end{aligned} ]

Bubble Pressure and the Laplace Term

A gas bubble in a solid has an internal pressure. For a spherical bubble, the pressure is increased by surface energy. The smaller the bubble, the more curved its surface is, and the larger this pressure term becomes:

[ P_\mathrm{gas} = P_\mathrm{ext} + \frac{2\gamma}{r} ]

This is the Young-Laplace equation. Here $\gamma$ is the surface energy and $r$ is the bubble radius. If external pressure is small compared with the surface term, the bubble pressure is approximately

[ P_\mathrm{gas} \approx \frac{2\gamma}{r} ]

Now use the ideal gas law:

[ PV=nRT ]

yielding

[ V_\mathrm{gas} = \frac{nRT}{P} = \frac{nRT}{P_\mathrm{ext}+2\gamma/r} ]

This equation shows the key behavior.

If the same number of gas atoms is stored in smaller bubbles, the pressure is higher and the gas occupies less volume. If bubbles coarsen and become larger, the pressure drops and the same gas inventory occupies more volume. Larger bubbles therefore cause more swelling.

Worked Swelling Example

We assume

[ N_f=10^{21}, \qquad Y_\mathrm{Xe}=0.065, \qquad T=800\mathrm{K} ] [ \gamma=1.0 \mathrm{J\,m^{-2}}, \qquad V_\mathrm{fuel}=0.40 \mathrm{cm^3} ]

We also neglect external pressure compared with the Laplace pressure. For a $5\mathrm{nm}$ bubble radius,

[ \begin{aligned} P &= \frac{2\gamma}{r} \\ &= \frac{2(1.0)}{5\times10^{-9}} \\ &= 4.0\times10^8 \mathrm{Pa} \end{aligned} ]

The gas volume is

[ V_\mathrm{gas} = \frac{(1.08\times10^{-4})(8.314)(800)}{4.0\times10^8} = 1.79\times10^{-9} \mathrm{m^3} ]

This is

[ V_\mathrm{gas}=1.79\times10^{-3}\mathrm{cm^3} ]

The fractional swelling is

[ \frac{\Delta V}{V_\mathrm{fuel}} = \frac{1.79\times10^{-3}}{0.40} =4.48\times10^{-3} =0.45\% ]

For $20\mathrm{nm}$ bubbles, the radius is four times larger. The pressure is four times lower, so the gas volume and swelling are four times larger:

[ \left(\frac{\Delta V}{V}\right)_{20\,\mathrm{nm}} \approx 1.79\% ]

For $50\mathrm{nm}$ bubbles:

[ \left(\frac{\Delta V}{V}\right)_{50\,\mathrm{nm}} \approx 4.49\% ]

The exact numerical result depends on the assumed fuel volume, temperature, surface energy and whether external pressure is neglected. In addition it was assumed that all Xeon stays trapped. The trend is: larger bubbles produce more swelling for the same gas inventory.

Radioactive Decay

Radioactive decay is the spontaneous transformation of unstable nuclei. For a single isotope, the number of atoms $N$ decreases according to

[ \frac{\mathrm{d}N}{\mathrm{d}t} = -\lambda N ]

The solution is

[ N(t)=N_0 e^{-\lambda t} ]

The decay constant $\lambda$ is related to half life by

[ \lambda = \frac{\ln 2}{T_{1/2}} ]

Activity is the number of decays per second:

[ A=\lambda N ]

The SI unit is the becquerel (Bq):

[ 1\mathrm{Bq}=1\mathrm{decay/s} ]

For a single isolated isotope, activity follows the same exponential law:

[ \begin{aligned} A(t) &=A_0e^{-\lambda t} \\ &=A_0 2^{-t/T_{1/2}} \end{aligned} ]

For example, if an isotope has initial activity of $A_0=7843\mathrm{Bq}$ and a half-life time of $T_{1/2}=7.1\mathrm{h}$, then after $t=10\mathrm{h}$ we get

[ A(10\mathrm{h}) = 7843\cdot2^{-10/7.1} =2955 \mathrm{Bq} ]

A short half-life means a large decay constant. For the same number of atoms, a short-lived isotope has a higher activity than a long-lived isotope.

Activation

Activation means that a nuclear reaction creates a radioactive product. A classic example is cobalt activation:

[ {}^{59}\mathrm{Co}(n,\gamma){}^{60}\mathrm{Co} ]

The neutron is captured and a gamma ray is emitted. The product, cobalt-60, is radioactive. The reaction rate per target atom is

[ r=\phi\sigma ]

where $\phi$ is flux and $\sigma$ is the microscopic cross section. If there are $N$ target atoms, the total production rate is

[ R=\phi\sigma N ]

If the product nucleus also decays while it is being produced, then

[ \frac{\mathrm{d}N^*}{\mathrm{d}t} = \phi\sigma N - \lambda N^* ]

product inventory increases by production and decreases by decay. If the target inventory $N$ is approximately constant, the solution is

[ N^*(t) = \frac{\phi\sigma N}{\lambda} \left(1-e^{-\lambda t}\right) ]

The activity is

[ A(t) = \lambda N^*(t) = \phi\sigma N \left(1-e^{-\lambda t}\right) ]

For very long irradiation times compared with the half life time, the exponential term becomes small and the activity approaches saturation:

[ A_\mathrm{sat}=\phi\sigma N ]

After irradiation stops, production stops. The product then simply decays:

[ A(t_\mathrm{cool}) = A_\mathrm{EOI}e^{-\lambda t_\mathrm{cool}} ]

where EOI means end of irradiation.

U-238 to Pu-239 Conversion

Uranium-238 is called fertile because it can become fissile plutonium-239 after neutron capture and beta decay:

[ {}^{238}\mathrm{U}(n,\gamma){}^{239}\mathrm{U} \xrightarrow{\beta^-} {}^{239}\mathrm{Np} \xrightarrow{\beta^-} {}^{239}\mathrm{Pu} ]

A simple first calculation asks only what fraction of U-238 atoms capture a neutron in a given time. The capture rate per U-238 atom is

[ k=\phi\sigma ]

The fraction transformed by neutron capture after time $t$ is

[ f =1-e^{-\phi\sigma t} ]

If $\phi\sigma t\ll1$, this becomes

[ f\approx\phi\sigma t ]

For example, take $\phi=10^{14} \mathrm{n\,cm^{-2}\,s^{-1}}$, $\sigma=2.7\mathrm{barn}=2.7\times10^{-24}\mathrm{cm^2}$ and $t=1\mathrm{yr}=3.156\times10^7\mathrm{s}$. Then

[ \phi\sigma t = (10^{14})(2.7\times10^{-24})(3.156\times10^7) = 8.52\times10^{-3} ]

The exact exponential result is

[ f =1-e^{-0.00852} =0.00848 ]

or about 0.85%.

This calculation only counts the first neutron capture. A full reactor depletion calculation must also include neutron spectrum, changing isotopic composition, fission of Pu-239, further captures, leakage and power history. But the first-order result is useful because it shows why the product $\phi\sigma t$ is the natural quantity. Flux times cross section gives a reaction probability per second, and multiplying by time gives the accumulated probability.

Decay Chains and Bateman Equations

Activation and transmutation rarely produce just one isolated isotope. Usually there are chains. The simplest chain is

[ A\rightarrow B ]

The parent isotope $A$ decays with decay constant $\lambda_A$:

[ \frac{\mathrm{d}N_A}{\mathrm{d}t} = -\lambda_A N_A ]

The daughter isotope $B$ is produced by decay of $A$ and removed by its own decay:

[ \frac{\mathrm{d}N_B}{\mathrm{d}t} = \lambda_A N_A-\lambda_B N_B ]

If initially $N_A(0)=N_{A0}$ and $N_B(0)=0$, then

[ \begin{aligned} N_A(t) &= N_{A0}e^{-\lambda_A t} \\ N_B(t) &= \frac{\lambda_A N_{A0}}{\lambda_B-\lambda_A} \left(e^{-\lambda_A t}-e^{-\lambda_B t}\right) \end{aligned} ]

These are the simplest Bateman equations. Real activation and depletion codes solve many coupled equations of the same type. In matrix form one may write

[ \frac{\mathrm{d}\mathbf{N}}{\mathrm{d}t} = \mathbf{A}\mathbf{N} ]

where $\mathbf{N}$ contains isotope inventories and $\mathbf{A}$ contains decay constants, neutron reaction rates, branching ratios and fission yields.

This is why activation codes need nuclear data libraries. The equations are conceptually simple, but the data set is large.

What the Common Tools are for

SRIM is useful for ion irradiation. It estimates stopping, ion range, recoil production and vacancy production versus depth. It is well suited for converting ion current and irradiation time into an approximate damage profile. It should not be used by itself to predict long-term swelling, hardening or retained defect fractions.

QCalc is useful for educational decay and activation estimates. It is a good way to check half-lives, exponential decay, simple buildup and simple decay chains.

FISPACT-II is an activation and inventory code. It is used for activation, transmutation, decay heat, dose-rate source terms and gas production using evaluated nuclear data libraries.

ORIGEN is a depletion and decay code widely used for reactor fuel inventory, spent fuel composition, decay heat and source terms.

OpenMC is a neutron transport code that can also perform depletion calculations. It is useful when geometry and neutron spectrum matter, because reaction rates come from the transport calculation rather than from a single assumed flux.

The hand calculations in these notes are best viewed as sanity checks. If a code result differs by six orders of magnitude from a hand estimate, the first suspect is usually a unit error, spectrum mismatch or wrong material inventory.

Calculation Checklist

For dpa from SRIM:

• Read the damage value in vacancies per ion per Angstrom.
• Convert current to ions per second.
• Multiply by irradiation time and divide by area to get fluence.
• Convert atom density to atoms per square centimeter per Angstrom.
• Divide vacancies produced by atoms present.

For fission gas swelling:

• Compute gas atoms from fissions and yield.
• Convert atoms to moles.
• Compute bubble pressure with $P=P_\mathrm{ext}+2\gamma/r$.
• Compute gas volume with $PV=nRT$.
• Divide gas volume by fuel volume.

For activation:

• Convert cross section to square centimeters.
• Compute production rate $R=\phi\sigma N$.
• Include decay during irradiation with $1-e^{-\lambda t}$.
• Include cooling after irradiation with $e^{-\lambda t_\mathrm{cool}}$.

For decay chains:

• Write the chain explicitly.
• Convert all half-lives to decay constants.
• Use Bateman equations for small chains.
• Use an activation or depletion code for realistic isotope networks.

References

• Olander, D. R. (1975). Fundamental aspects of nuclear reactor fuel elements (No. TID-26711-P1). California Univ., Berkeley (USA). Dept. of Nuclear Engineering.
• Was, G. S. (2007). Fundamentals of radiation materials science: metals and alloys. Berlin, Heidelberg: Springer Berlin Heidelberg.
• Murty, K. L., & Charit, I. (2013). An introduction to nuclear materials: fundamentals and applications. John Wiley & Sons.
• University lectures at Technical University of Vienna:
  • 308.895 Materialien in nuklearen Anwendungen – Grundlagen
  • 308.150 Materialien in nuklearen Anwendungen – Vertiefende Themen
• Tools:
  • SRIM
  • QCalc
  • FISPACT-II
  • ORIGEN
  • OpenMC

This article is tagged: Physics, Basics, Tutorial, Mechanics, Nuclear