11.8: Artificial Transmutation
- Page ID
- 86621
- Describe transmutation.
- Write and balance transmutation equations.
Although the conversion of one element to another is the basis of natural radioactive decay, it is also possible to convert one element to another artificially. The conversion of one element to another is the process of transmutation. Between 1921 and 1924, Patrick Blackett conducted experiments in which he converted a stable isotope of nitrogen to a stable isotope of oxygen. By bombarding \(\ce{^{14}N}\) with \(\alpha\) particles he created \(\ce{^{17}O}\). Transmutation may also be accomplished by bombardment with neutrons.
\[\ce{^{14}_7N + ^4_2He \rightarrow ^{17}_8O + ^1_1H} \nonumber\]
The \(\ce{^{17}_8O}\) and \(\ce{^1_1H}\) nuclei that are produced are stable, so no further (nuclear) changes occur.
To reach the kinetic energies necessary to produce transmutation reactions, devices called particle accelerators are used. These devices use magnetic and electric fields to increase the speeds of nuclear particles. In all accelerators, the particles move in a vacuum to avoid collisions with gas molecules. When neutrons are required for transmutation reactions, they are usually obtained from radioactive decay reactions or from various nuclear reactions occurring in nuclear reactors.
Located near Geneva, the CERN (“Conseil Européen pour la Recherche Nucléaire,” or European Council for Nuclear Research) Laboratory is the world’s premier center for the investigations of the fundamental particles that make up matter. It contains the 27-kilometer (17 mile) long, circular Large Hadron Collider (LHC), the largest particle accelerator in the world (Figure \(\PageIndex{1}\)). In the LHC, particles are boosted to high energies and are then made to collide with each other or with stationary targets at nearly the speed of light. Superconducting electromagnets are used to produce a strong magnetic field that guides the particles around the ring. Specialized, purpose-built detectors observe and record the results of these collisions, which are then analyzed by CERN scientists using powerful computers.
Figure \(\PageIndex{1}\): A small section of the LHC is shown with workers traveling along it. (credit: Christophe Delaere)
In 2012, CERN announced that experiments at the LHC showed the first observations of the Higgs boson, an elementary particle that helps explain the origin of mass in fundamental particles. This long-anticipated discovery made worldwide news and resulted in the awarding of the 2103 Nobel Prize in Physics to François Englert and Peter Higgs, who had predicted the existence of this particle almost 50 years previously.
Prior to 1940, the heaviest-known element was uranium, whose atomic number is 92. Now, many artificial elements have been synthesized and isolated, including several on such a large scale that they have had a profound effect on society. One of these—element 93, neptunium (Np)—was first made in 1940 by McMillan and Abelson by bombarding uranium-238 with neutrons. The reaction creates unstable uranium-239, with a half-life of 23.5 minutes, which then decays into neptunium-239. Neptunium-239 is also radioactive, with a half-life of 2.36 days, and it decays into plutonium-239. The nuclear reactions are:
\ce{^{238}_{92}U + ^1_0n} &⟶ \ce{^{239}_{92}U} && \\
\ce{^{239}_{92}U} &⟶ \ce{^{239}_{93}Np + ^0_{−1}e} &&\textrm{half-life}=\mathrm{23.5\: min} \\
\ce{^{239}_{93}Np } &⟶ \ce{^{239}_{94}Pu + ^0_{−1}e} &&\textrm{half-life}=\mathrm{2.36\: days}
\end{align*}\]
Plutonium is now mostly formed in nuclear reactors as a byproduct during the decay of uranium. Some of the neutrons that are released during U-235 decay combine with U-238 nuclei to form uranium-239; this undergoes β decay to form neptunium-239, which in turn undergoes β decay to form plutonium-239 as illustrated in the preceding three equations. It is possible to summarize these equations as:
\[\mathrm{\ce{^{238}_{92}U} + {^1_0n}⟶ \ce{^{239}_{92}U} \xrightarrow{β^-} \ce{^{239}_{93}Np} \xrightarrow{β^-} \ce{^{239}_{94}Pu}}\]
Heavier isotopes of plutonium—Pu-240, Pu-241, and Pu-242—are also produced when lighter plutonium nuclei capture neutrons. Some of this highly radioactive plutonium is used to produce military weapons, and the rest presents a serious storage problem because they have half-lives from thousands to hundreds of thousands of years.
Although they have not been prepared in the same quantity as plutonium, many other synthetic nuclei have been produced. Nuclear medicine has developed from the ability to convert atoms of one type into other types of atoms. Radioactive isotopes of several dozen elements are currently used for medical applications. The radiation produced by their decay is used to image or treat various organs or portions of the body, among other uses.
The elements beyond element 92 (uranium) are called transuranium elements. As of this writing, 22 transuranium elements have been produced and officially recognized by IUPAC; several other elements have formation claims that are waiting for approval. Some of these elements are shown in Table \(\PageIndex{1}\).
| Name | Symbol | Atomic Number | Reaction |
|---|---|---|---|
| americium | Am | 95 | \(\ce{^{239}_{94}Pu + ^1_0n ⟶ ^{240}_{95}Am + ^0_{−1}e}\) |
| curium | Cm | 96 | \(\ce{^{239}_{94}Pu + ^4_2He ⟶ ^{242}_{96}Cm + ^1_0n}\) |
| californium | Cf | 98 | \(\ce{^{242}_{96}Cm + ^4_2He⟶ ^{243}_{97}Bk + 2^1_0n}\) |
| einsteinium | Es | 99 | \(\ce{^{238}_{92}U + 15^1_0n⟶ ^{253}_{99}Es + 7^0_{−1}e}\) |
| mendelevium | Md | 101 | \(\ce{^{253}_{99}Es + ^4_2He ⟶ ^{256}_{101}Md + ^1_0n}\) |
| nobelium | No | 102 | \(\ce{^{246}_{96}Cm + ^{12}_6C ⟶ ^{254}_{102}No + 4 ^1_0n}\) |
| rutherfordium | Rf | 104 | \(\ce{^{249}_{98}Cf + ^{12}_6C⟶ ^{257}_{104}Rf + 4 ^1_0n}\) |
|
seaborgium |
Sg |
106 |
\(\ce{^{206}_{82}Pb + ^{54}_{24}Cr ⟶ ^{257}_{106}Sg + 3 ^1_0n}\) \(\ce{^{249}_{98}Cf + ^{18}_8O ⟶ ^{263}_{106}Sg + 4 ^1_0n}\) |
| meitnerium | Mt | 107 | \(\ce{^{209}_{83}Bi + ^{58}_{26}Fe ⟶ ^{266}_{109}Mt + ^1_0n}\) |
Write the balanced nuclear equation for the production of the following transuranium elements:
- berkelium-244, made by the reaction of Am-241 and He-4
- fermium-254, made by the reaction of Pu-239 with a large number of neutrons
- lawrencium-257, made by the reaction of Cf-250 and B-11
- dubnium-260, made by the reaction of Cf-249 and N-15
Solution
a
From the given information we can write the nuclear equation
\[ ^{241}_{95}\text{Am} +^4_2\text{He}\rightarrow ^{244}_{97}\text{Bk} \nonumber\]
On the left side the total mass number is
\[ 241 + 4 = 245 \nonumber\]
and the total atomic number is
\[ 95 + 2 = 97 \nonumber\]
On the right side the total mass number is 244 and the total atomic number is 97. This shows that one neutron needs to be added which would increase the total mass number needs to by one while keeping the total atomic number the same. The balanced nuclear equation would be
\[ ^{241}_{95}\text{Am} +^4_2\text{He}\rightarrow ^{244}_{97}\text{Bk}+^1_0\text{n} \nonumber\]
b
From the given information we can write the nuclear equation
\[ ^{239}_{94}\text{Pu}+\text{x }^1_0\text{n}\rightarrow^{254}_{100}\text{Fm} \nonumber\]
On the left side we see that the total mass number is the sum of \( 239+(1)\text{x}\). on the right side we see that the total mass number is 254. since the total mass number of the reactants must equal that of the products we can write
\[ 239+\text{x}=254 \nonumber\]
showing 15 neutrons need to be added to balance the mass number.
\[ \ce{^{239}_{94}Pu + 15 ^1_0n \rightarrow ^{254}_{100}Fm} \nonumber\]
To balance the total atomic number of the equation, 6 electrons need to be added to the right side. Therefore the balanced equation reads:
\[ ^{239}_{94}\text{Pu}+15\text{ }^1_0\text{n}\rightarrow^{254}_{100}\text{Fm}+\ce{6 ^0_{-1}e} \nonumber\]
c
From the given information we can write the nuclear equation
\[ ^{250}_{98}\text{Cf}+\ce{^{11}_5B}\rightarrow^{257}_{103}\text{Lr} \nonumber\]
On the left side the total mass number is
\[250+11=261 \nonumber\]
and the total atomic number is
\[98+5=103 \nonumber\]
On the right side the total mass number is 257 and the total atomic number is 103. This means that 4 neutrons need to be added to the right side to balance the equation. The balanced nuclear equation is
\[ ^{250}_{98}\text{Cf}+\ce{^{11}_5B}\rightarrow^{257}_{103}\text{Lr}+4\text{ }^1_0\text{n} \nonumber\]
d
From the given information we can write the nuclear equation
\[ ^{249}_{98}\text{Cf}+\ce{^{15}_7N}\rightarrow^{260}_{105}\text{Db} \nonumber\]
On the left side the total mass number is
\[249+15=264 \nonumber\]
and the total atomic number is
\[ 98+7=105 \nonumber\]
On the right side the total mass number is 260 and the total atomic number is 105. This means that 4 neutrons need to be added to the right side t obalance the equation. The balanced nuclear equation is
\[ ^{249}_{98}\text{Cf}+\ce{^{15}_7N}\rightarrow^{260}_{105}\text{Lr}+4\text{ }^1_0\text{n} \nonumber\]
Contributors and Attributions
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Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/85abf193-2bd...a7ac8df6@9.110).