Artificial Elements and Radionuclide Production

Introduction

Technetium, promethium and every element beyond uranium are essentially absent from the Earth's crust, yet chemists have studied them in detail. They were made artificially. Today, reactors and accelerators produce hundreds of radionuclides for medicine, industry and research, and heavy-ion facilities have extended the periodic table to oganesson, element 118. This page explains the two main production routes, how yields are estimated, and why the newest elements appear only a few atoms at a time.

Core explanation

Two families of production route. Radionuclides are made either by adding neutrons or by adding charged particles.

Reactor production. A research reactor provides an intense neutron flux (often 10¹³–10¹⁴ neutrons cm⁻² s⁻¹). Placing a target in the flux leads to (n,γ) capture, which adds one neutron and usually gives a neutron-rich nuclide that decays by beta-minus emission. Examples include cobalt-60 from cobalt-59 and molybdenum-99, which is also obtained as a fission product of uranium-235. Because (n,γ) products have the same Z as the target, they cannot be chemically separated from it, so the product has low specific activity unless a following beta decay changes the element.

Accelerator production. A cyclotron accelerates protons, deuterons or alpha particles to energies of several to tens of MeV. Charged particles must overcome the Coulomb barrier, so they need these high energies. Reactions such as (p,n) change Z, so the product is a different element and can be separated chemically to give carrier-free , high-specific-activity material. Cyclotron products are usually proton-rich and decay by positron emission or electron capture — ideal for PET imaging. Fluorine-18 is made by the ¹⁸O(p,n)¹⁸F reaction on oxygen-18-enriched water.

Yield during irradiation. The production rate R = Nσφ, where N is the number of target nuclei, σ the cross-section and φ the flux. The product decays while it forms, so its activity builds up as A = R(1 − e^(−λt)). After about five half-lives the activity approaches the saturation activity , equal to R. Irradiating for much longer gains almost nothing.

Transuranium elements by neutron capture. Neptunium (Z = 93) and plutonium (Z = 94) were first identified in 1940. Uranium-238 captures a neutron to form uranium-239, which beta-decays to neptunium-239 and then to plutonium-239. Repeated neutron capture and beta decay in high-flux reactors climb as far as fermium (Z = 100). Beyond this, very short-lived isotopes (such as those of fermium near mass 258, which undergo rapid spontaneous fission) block the path.

Heavier elements by heavy-ion fusion. Elements beyond fermium are made by firing medium-mass ions at heavy targets. Cold fusion reactions use lead or bismuth targets with ions such as iron or zinc, forming a compound nucleus with low excitation that loses one neutron. Hot fusion reactions use actinide targets with calcium-48 beams, giving more neutron-rich products. Cross-sections fall to picobarns or below, so a campaign of weeks may identify only a handful of atoms, recognised by their characteristic alpha-decay chains leading to known nuclides.

Formulae

Production rate: R = Nσφ.

Activity after irradiation time t: A = R(1 − e^(−λt)); saturation activity A sat = R.

Decay after the end of irradiation: A(t') = A₀e^(−λt').

Step-by-step reasoning

To choose a production route for a radionuclide:

1. Decide whether the target nuclide is neutron-rich (reactor) or proton-rich (cyclotron). 2. Identify a target and a reaction with a suitable cross-section. 3. Check whether the product is a different element, so it can be separated carrier-free. 4. Choose an irradiation time of a few half-lives to approach saturation. 5. Allow for decay during chemical processing and transport before use.

Visual explanation

On the chart of nuclides, an (n,γ) reaction moves one square to the right (more neutrons) at the same Z, and beta-minus decay moves one square diagonally up and to the left. A (p,n) reaction moves diagonally up one row and one column left. Heavy-ion fusion jumps a long way up and to the right in a single step, landing far into the superheavy region.

Real-world analogy

Reactor production is like soaking a sponge in a steady drizzle: every target nucleus is exposed to the neutron rain and slowly absorbs. Cyclotron production is like a firing range: a narrow beam of fast projectiles strikes a small target, and only direct hits with enough energy produce a change.

Real-world example

Most technetium-99m used in hospitals comes from molybdenum-99, which is made mainly by fission of uranium-235 in research reactors. A handful of reactors supply much of the world's demand, so reactor shutdowns have caused international shortages, which is one reason many countries now invest in cyclotron-based alternatives.

Why?

Why is an irradiation longer than a few half-lives wasteful? The rate of formation is constant, but the rate of decay rises in proportion to the number of product nuclei present. After five half-lives decay almost exactly balances formation, so activity stays at about 97% of saturation however much longer the target is irradiated.

Common misconception

"Bombarding a target for twice as long always gives twice the activity." This is true only for irradiation times much shorter than the half-life. Activity approaches a saturation value, so doubling a long irradiation barely changes the yield.

Worked example

Question: A target is irradiated for one half-life of the product. What fraction of the saturation activity is reached? What fraction after three half-lives?

Reasoning: A/A sat = 1 − e^(−λt). After one half-life, e^(−λt) = 1/2, so A/A sat = 0.50. After three half-lives, e^(−λt) = 1/8, so A/A sat = 1 − 0.125 = 0.875.

Answer: 50% after one half-life; 87.5% after three half-lives.

Quick check

1. Why can cyclotron-produced fluorine-18 be obtained with very high specific activity? Answer: The (p,n) reaction changes oxygen into fluorine, a different element, so fluorine-18 can be separated chemically from the target without stable carrier.

Exam focus

Examiners often ask you to complete production equations, such as ²³⁸U(n,γ)²³⁹U followed by two beta decays to ²³⁹Pu, and to explain why reactor products tend to be beta-minus emitters while cyclotron products tend to be positron emitters or decay by electron capture. Know the saturation formula and its shape.

Advanced insight

The discovery of superheavy elements depends on correlation rather than chemistry in the usual sense: an implanted atom is identified by a sequence of alpha decays with specific energies and times, ending in known nuclides. Remarkably, single-atom chemistry has been carried out on elements such as seaborgium and flerovium by gas-phase adsorption, testing whether relativistic effects on inner electrons change their expected periodic-table behaviour.

Summary

Radionuclides are made by neutron irradiation in reactors, which gives neutron-rich beta-minus emitters, and by charged-particle bombardment in cyclotrons, which gives proton-rich positron emitters that can often be separated carrier-free. Yield follows A = R(1 − e^(−λt)) and saturates after a few half-lives. Neutron capture builds elements up to fermium; heavier elements require heavy-ion fusion with tiny cross-sections.

Practice questions

1. Write the sequence of reactions that makes plutonium-239 from uranium-238. Answer: ²³⁸U captures a neutron to form ²³⁹U, which beta-decays to ²³⁹Np, which beta-decays to ²³⁹Pu. 2. Why do charged particles need much higher energies than neutrons to cause nuclear reactions? Answer: They must overcome or tunnel through the Coulomb repulsion of the target nucleus, whereas neutrons are uncharged and feel no such barrier. 3. A product has a half-life of 6 hours. Estimate a sensible irradiation time and explain your choice. Answer: About 12–30 hours (two to five half-lives), because activity then approaches saturation and further irradiation adds very little. 4. Explain why the (n,γ) product of a stable target usually has low specific activity. Answer: The product is the same element as the target, so it cannot be separated chemically and is diluted by a large excess of the stable isotope.