Isotope Production and Burnup Calculator

Compute isotope production activity and saturation fraction from neutron flux, activation cross-section, and irradiation time, or find target burnup fraction and remaining atoms.

⚛️ Isotope Production and Burnup Calculator
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Activity A(t)
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Saturation Activity Asat
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Saturation Fraction
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Decay Constant λ
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Fraction Burned
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Remaining Atoms N(t)
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Initial Reaction Rate R0
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Current Reaction Rate R(t)
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⚛️ What Is Isotope Production and Target Burnup?

Isotope production by neutron activation is the process of placing a stable target material in a neutron flux, causing target nuclei to absorb neutrons and transmute into radioactive product isotopes. The activity of the product builds up over time according to A(t) = N × σ × φ × (1 − exp(−λt)), where N is the number of target atoms, σ is the neutron activation cross-section in barns (1 barn = 10−24 cm2), φ is the neutron flux in n/cm2/s, and λ = ln(2)/T1/2 is the decay constant of the radioactive product. This formula accounts for both the continuous production of new product atoms by neutron capture and their simultaneous radioactive decay.

The quantity N × σ × φ is the saturation activity Asat, the maximum activity theoretically achievable by irradiating the target at this flux for an infinite time. The factor (1 − exp(−λt)) is the saturation fraction, which rises from 0 at t = 0 and approaches 1 asymptotically. After one product half-life, the saturation fraction is exactly 50%. After 3.32 half-lives, it reaches 90%. This diminishing-return relationship shapes the economics of isotope production: irradiating Mo-99 (T1/2 = 66 hr) targets for 5 to 7 days reaches 85 to 94% of saturation, while irradiating for 2 weeks adds only a small additional yield.

Production mode is used across nuclear medicine (Mo-99/Tc-99m generators, I-131, Lu-177, Y-90), industrial radiography (Ir-192, Co-60, Se-75), environmental tracing (Au-198 sediment tracers), and analytical chemistry (neutron activation analysis). Burnup mode covers the complementary problem: tracking the depletion of a target material, such as a boron control poison, a uranium fuel nuclide, or a neutron absorber used in shielding, where the quantity of interest is how much of the original material remains after a given irradiation period at a given flux.

Both calculations share the same mathematical structure: an exponential approach to a limiting value. In production, the approach is to saturation activity controlled by the product decay constant. In burnup, the approach is to complete depletion controlled by the effective burnup rate constant σφ. This calculator covers both modes with scientific-notation input for the flux and atom count, since these quantities routinely span 15 orders of magnitude across different applications.

📐 Formulas

Production: A(t) = N × σ × φ × (1 − e−λt)
Saturation Activity: Asat = N × σ × φ
Burnup: B(t) = 1 − e−σφt    N(t) = N0 × e−σφt
A(t) = activity of product isotope at time t (Bq)
N = number of target atoms (dimensionless count)
σ = neutron activation or absorption cross-section (barns; 1 b = 10−24 cm2)
φ = neutron flux (n/cm2/s)
λ = decay constant of product = ln(2) / T1/2 (s−1)
t = irradiation time (s, converted from the selected unit)
B(t) = fraction of target atoms consumed by neutron absorption (dimensionless, 0 to 1)
N0 = initial target atom count; N(t) = remaining atoms at time t
Example (Au-197): N = 1020, σ = 98.7 b, φ = 1013 n/cm2/s, T1/2 = 2.695 day, t = 3 day: A(t) = 53.1 GBq, Asat = 98.7 GBq, saturation fraction = 53.8%

📖 How to Use This Calculator

Steps

1
Select calculation mode - choose "Isotope Production" to compute activity buildup during irradiation, or "Target Burnup" to compute the fraction of target atoms consumed by neutron absorption and the remaining atom count.
2
Enter neutron flux - type the mantissa and exponent of the thermal neutron flux. For example, a research reactor flux of 3.0 × 1013 n/cm2/s is entered as mantissa 3.0 and exponent 13. Consult your reactor operator for measured flux values at the sample position.
3
Enter cross-section and target atoms - look up the activation or absorption cross-section for your target nuclide from NNDC or ENDF/B nuclear data libraries. Enter the number of target atoms using mantissa and exponent fields. To convert mass to atoms: N = (m × 6.022×1023) / A, where m is mass in grams and A is atomic mass in g/mol.
4
Enter half-life and irradiation time (Production mode) - type the product isotope half-life and select its time unit. Then enter the planned irradiation time. The defaults are pre-loaded with Au-197 activation data for a 3-day irradiation.
5
Read the results - Production mode shows activity at time t, saturation activity (upper limit), saturation fraction, and decay constant. Burnup mode shows fraction burned, remaining atoms, and initial versus current reaction rate. Use "Copy link" to save and share your specific inputs.

💡 Example Calculations

Example 1 - Au-197 Neutron Activation (3-Day Irradiation)

Au-197 target: 1020 atoms, σ = 98.7 barns, φ = 1013 n/cm2/s, T1/2(Au-198) = 2.695 days, t = 3 days

1
Compute saturation activity: Asat = N × σ × φ = 1020 × 98.7×10−24 cm2 × 1013 = 98.70 GBq.
2
Compute λ: λ = ln(2) / (2.695 × 86400 s) = 2.977×10−6 s−1. Then λt = 2.977×10−6 × 259200 = 0.7717.
3
Saturation fraction = 1 − exp(−0.7717) = 0.5381 = 53.81%. Activity = 98.70 × 0.5381 = 53.10 GBq.
A(3 days) = 53.10 GBq  |  Asat = 98.70 GBq  |  Saturation = 53.81%
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Example 2 - Co-60 Production (One-Year Reactor Irradiation)

Co-59 target: 6×1021 atoms, σ = 37.2 barns, φ = 5×1013 n/cm2/s, T1/2(Co-60) = 5.27 yr, t = 365 days

1
Asat = 6×1021 × 37.2×10−24 × 5×1013 = 11.16 TBq. This is the ceiling activity at infinite irradiation time.
2
λ(Co-60) = ln(2) / (5.27 × 31,557,600 s) = 4.168×10−9 s−1. λt = 4.168×10−9 × 31,536,000 = 0.1314.
3
Saturation fraction = 1 − exp(−0.1314) = 12.31%. Activity = 11.16 × 0.1231 = 1.374 TBq per kilogram of cobalt metal in the target assembly.
A(365 days) = 1.374 TBq  |  Asat = 11.16 TBq  |  Saturation = 12.31%
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Example 3 - B-10 Control Poison Burnup (30-Day Irradiation)

B-10 absorber: N0 = 1022 atoms, σ = 3840 barns, φ = 1014 n/cm2/s, t = 30 days

1
Effective burnup constant: σφ = 3840×10−24 × 1014 = 3.84×10−7 s−1. Over t = 30 days = 2.592×106 s: σφt = 0.9953.
2
Fraction burned = 1 − exp(−0.9953) = 63.01%. Remaining atoms: N(t) = 1022 × exp(−0.9953) = 3.699×1021 atoms.
3
Initial reaction rate R0 = 1022 × 3.84×10−7 = 3.84×1015 rxn/s. Current rate at t = 30 days: R(t) = 3.699×1021 × 3.84×10−7 = 1.420×1015 rxn/s.
Burned = 63.01%  |  Remaining = 3.70×1021 atoms  |  R0 = 3.84×1015 rxn/s
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❓ Frequently Asked Questions

What is the isotope production formula A(t) = N·σ·φ·(1 - exp(-λt))?+
This formula gives the radioactivity of a product isotope at time t during neutron irradiation. N is the number of target atoms, σ is the activation cross-section in cm² (1 barn = 10^-24 cm²), φ is the neutron flux in n/cm²/s, and λ = ln(2)/T1/2 is the decay constant of the product. The term (1 - exp(-λt)) is the saturation fraction, ranging from 0 at t=0 to 1 as t grows much larger than T1/2. At t = T1/2, the saturation fraction is exactly 0.5.
What is saturation activity and why does it matter for isotope production?+
Saturation activity A_sat = N·σ·φ is the maximum possible activity from a given target at a given flux. It equals the rate of production of the radioactive product at time zero, before any decay has occurred. At saturation, the production rate exactly equals the decay rate and the activity stops increasing. Saturation activity sets the ceiling for isotope production: to increase it you must increase the flux, the target mass, or use a target material with a larger cross-section.
How long does it take to reach 90% of saturation activity?+
Setting the saturation fraction to 0.90 gives 1 - exp(-λt) = 0.90, so λt = ln(10) = 2.303, and t = 2.303/λ = 2.303 × T1/2 / ln(2) = 3.32 × T1/2. To reach 99% of saturation requires 6.64 half-lives. For a short-lived isotope like Mo-99 (T1/2 = 65.94 hr), 90% saturation takes about 9.2 days of continuous irradiation.
What is neutron burnup and how does it differ from isotope production?+
Neutron burnup tracks the depletion of a target nuclide by neutron absorption without tracking radioactive ingrowth of a product. The burnup fraction B = 1 - exp(-σ·φ·t) measures what fraction of the original target atoms have been consumed. Burnup mode is appropriate when the product is stable or when you are only interested in fuel or poison depletion. Production mode is used when you care about the radioactivity that builds up in the product isotope.
What neutron flux is needed to produce a specific activity of a medical isotope?+
Rearranging A_sat = N·σ·φ gives the required flux: φ = A_sat / (N·σ). For Au-198 production with 1 g of Au-197 (N = 3.07×10^21 atoms), σ = 98.7 barns = 9.87×10^-23 cm², and a target saturation activity of 10 GBq: φ = 10^10 / (3.07×10^21 × 9.87×10^-23) = 3.3×10^10 n/cm²/s. This is achievable in a low-flux research reactor. High-specific-activity medical isotopes require fluxes of 10^13 to 10^14 n/cm²/s.
How is Co-60 produced for industrial irradiators and cancer therapy units?+
Co-59 (natural cobalt) is loaded into a reactor and irradiated for 18 months to 3 years at a thermal neutron flux of about 3×10^13 n/cm²/s. The activation cross-section of Co-59 is 37.2 barns. Co-60 has a half-life of 5.27 years. Typical irradiation produces 10 to 20% of saturation activity, yielding 50 to 100 TBq per kilogram of cobalt. The product is used in food irradiation, industrial gamma radiography, and Gamma Knife radiosurgery.
What is the activation cross-section and how does it differ from the absorption cross-section?+
The activation cross-section (radiative capture cross-section, σ_γ) quantifies the probability of neutron capture that produces a radioactive product via the (n,γ) reaction. The absorption cross-section σ_abs includes all reactions that remove a neutron: radiative capture, fission (for fissile nuclei), and other inelastic reactions. For non-fissile materials, σ_activation and σ_abs are nearly equal. For U-235, σ_abs = 683 b while σ_fission = 582 b and σ_capture = 101 b.
What is B-10 burnup and why is it important in nuclear reactor control?+
B-10 (20% of natural boron) has a thermal neutron absorption cross-section of 3840 barns, among the highest of any stable nuclide. Control rods and neutron poisons use B-10 to absorb neutrons and suppress reactivity. During reactor operation, B-10 is consumed at rate σ·φ·N. After 1000 hours at a flux of 10^14 n/cm²/s, the burnup fraction is about 26%. Tracking B-10 depletion is essential for predicting control rod worth over a fuel cycle.
How do I convert target mass to atom count N for this calculator?+
Use N = (m × N_A) / A, where m is mass in grams, N_A = 6.022×10^23 atoms/mol (Avogadro constant), and A is the nuclide atomic mass in g/mol (approximately the mass number). For 1 mg of pure Mo-98 (A = 97.91 g/mol): N = (0.001 × 6.022×10^23) / 97.91 = 6.15×10^18 atoms. Enter this as mantissa 6.15, exponent 18 in the calculator.
What neutron flux values are typical in different irradiation facilities?+
Research reactors (NIST, ILL): thermal flux 10^14 to 2×10^15 n/cm²/s. Power reactor fuel center: 3 to 5×10^13 n/cm²/s. Medical cyclotron neutron beam: 10^8 to 10^12 n/cm²/s. Am-Be or Cf-252 neutron sources: 10^5 to 10^8 n/cm²/s. High-flux material test reactors (HFIR): up to 2×10^15 n/cm²/s in the reflector. The flux directly scales the production rate and burnup rate for a given target and cross-section.
Why does burnup use 1 - exp(-σ·φ·t) instead of a radioactive decay equation?+
In burnup, target atoms are removed by neutron absorption, not by spontaneous radioactive decay. The removal rate is proportional to σ·φ·N(t), giving dN/dt = -σ·φ·N, whose solution is N(t) = N_0 × exp(-σ·φ·t). The product σ·φ plays the role of an effective decay constant. The key difference from radioactive decay is that σ·φ depends on reactor operating conditions and can be changed by adjusting the flux, whereas the radioactive decay constant λ is fixed by nuclear physics.
What is the relationship between saturation fraction and the number of product half-lives elapsed?+
The saturation fraction S(t) = 1 - exp(-λt) depends only on the ratio t / T1/2. At t = T1/2, S = 0.500. At t = 2×T1/2, S = 0.750. At t = 3×T1/2, S = 0.875. Each additional half-life closes half of the remaining gap to saturation. This diminishing return explains why irradiating beyond 3 to 4 half-lives is rarely economical: you must double the irradiation time to close half the remaining gap to A_sat.
Tags: Isotope Production Calculator Neutron Activation Calculator Saturation Activity Formula Target Burnup Calculator Nuclear Irradiation Calculator Activation Cross-Section Calculator Formula Free Online

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📌 Quick Tips

💡The saturation activity A_sat = N·σ·φ is the maximum activity you can ever achieve by irradiating this target at this flux, no matter how long you irradiate. Longer irradiation simply approaches this limit asymptotically.
💡To reach 99% of saturation activity, irradiate for approximately 6.64 product half-lives (since 1 - exp(-6.64 × ln2) = 0.99). For Co-60 (T1/2 = 5.27 yr), this requires about 35 years, so industrial Co-60 targets are typically irradiated for 18 months to 2 years, reaching only 20-30% saturation.
💡For medical isotopes with short half-lives such as Mo-99 (66 hr), saturation is reached in about 14 days. Most commercial Mo-99 targets are irradiated for 5-7 days, reaching 70-90% of saturation activity.
💡Burnup fraction B = 1 - exp(-σ·φ·t) equals the saturation fraction formula with decay constant λ replaced by σ·φ. For a nuclide with no decay product of interest, use burnup mode to track depletion of the target material.
💡Atom count N relates to mass by N = (m × N_A) / A, where m is the target mass in grams, N_A = 6.022×10^23 atoms/mol, and A is the atomic mass in g/mol. For Au-197 (A = 197 g/mol), 1 mg of gold contains 3.07×10^18 atoms.