Deuterium

Hydrogen-2
Full table
General
symbol	deuterium, ²H or D
Neutrons	1
Protons	1
Nuclide Data
Natural abundance	0.015%
Half-life	stable
Isotope mass	2.01355321270 u
Spin	1+
Excess energy	13135.720 ± 0.001 keV
Binding energy	{{{binding_energy}}} ± {{{error2}}} keV

Deuterium, also called heavy hydrogen, is a Sun, since thermonuclear reactions destroy it. However, it continues to persist in the outer solar atmosphere at roughly the same concentration as in Jupiter.

The neutron, whereas the far more common hydrogen nucleus contains no neutrons. The isotope name is formed from the Greek deuteros meaning "second", to denote the two particles comprising the nucleus.^[2]

Differences between deuterium and common hydrogen (protium)

Chemical symbol

Deuterium is frequently represented by the atomic weight of 1.007947 u, and protium's mass of 1.007825 u. The isotope weight ratios within other chemical elements are largely insignificant in this regard, explaining the lack of unique isotope symbols elsewhere.^{[citation needed]}

Natural abundance

Deuterium occurs in trace amounts naturally as deuterium gas, written ²H₂ or D₂, but most natural occurrence in the universe is bonded with a typical ¹H atom, a gas called hydrogen deuteride (HD or ¹H²H).^[4]

The existence of deuterium on Earth, elsewhere in the solar system (as confirmed by planetary probes), and in the spectra of stars, is an important datum in cosmology. Stellar fusion destroys deuterium, and there are no known natural processes (for example, see the rare steady state theory of the universe.

The world's leading "producer" of deuterium (technically, merely enricher or concentrator of deuterium) was Canada, until 1997 when the last plant was shut down (see more in the neutron moderator for the operation of the CANDU reactor design. India is now probably the world's largest concentrator of heavy water, also used in nuclear power reactors.

Physical properties

The physical properties of deuterium compounds can be different from the hydrogen analogs; for example, D₂O is more H₂O.^{[citation needed]}

Deuterium behaves chemically similarly to ordinary hydrogen, but there are differences in bond energy and length for compounds of heavy hydrogen isotopes which are larger than the isotopic differences in any other element. Bonds involving deuterium and heavy water).

Deuterium can replace the normal hydrogen in water molecules to form health threat to humans unless very large quantities (in excess of 10 liters) were consumed over many days. Small doses of heavy water (a few grams in humans, containing an amount of deuterium comparable to that normally present in the body) are routinely used as harmless metabolic tracers in humans and animals.

Quantum properties

The deuteron has spin +1 and is thus a mass spectrometry.

Nuclear properties

Deuterium is one of only four stable nuclides with an odd number of protons and odd number of neutrons. (²H, ⁶Li, ¹⁰B, ¹⁴N; also, the long-lived radioactive nuclides ⁴⁰K, ⁵⁰V, ¹³⁸La, ^180mTa occur naturally.) Most odd-odd nuclei are unstable with respect to orbital angular momentum. But that would require orbital angular momentum and kinetic energy, so that they have a higher total energy (both due to their kinetic energy and because their distance would be larger and their binding energy lower). In both cases, this cases the di-proton and di-neutron nucleus to be unstable.

Deuterium as an isospin singlet

Due to the similarity in mass and nuclear properties between the neutron is known as isospin and denoted $τ$ .

Isospin is an SU(2) symmetry, like ordinary spin, so is completely analogous to it. The proton.

A pair of nucleons can either be in an antisymmetric state of isospin called singlet, or in a symmetric state called triplet. In terms of the "down" state and "up" state, the singlet is

$\frac{1}{\sqrt{2}}\Big( |\uparrow \downarrow \rangle - |\downarrow \uparrow \rangle\Big)$

This is a nucleus with one proton and one neutron, i.e. a deuterium nucleus.

The triplet is $\left( \begin{array}{ll} \uparrow\uparrow\\ \frac{1}{\sqrt{2}}(\uparrow\downarrow + \downarrow\uparrow)\\ \downarrow\downarrow \end{array} \right)$ And thus consists of three types of nuclei, which are supposed to be symmetric - a deuterium nucleus (actually a highly excited state of it), a nucleus with two neutrons. The latter two nuclei are not stable or nearly stable, and therefore so is this type of deuterium (meaning that it is indeed a highly excited state of deuterium).

Approximated wavefunction of the deuteron

The total wavefunction of both the nucleons: if it is even then the parity is even (positive), and if it is odd then the parity is odd (negative).

The deuteron, being an isospin singlet, is antisymmetric under nucleons exchange due to isospin, and therefore must be symmetric under the double exchange of their spin and location. Therefore it can be in either of the following two different states:

Symmetric spin and symmetric under parity. In this case, the exchange of the two nucleons will multiply the deuterium wavefunction by (-1) from isospin exchange, (+1) from spin exchange and (+1) from parity (location exchange), for a total of (-1) as needed for antisymmetry.
Antisymmetric spin and antisymmetric under parity. In this case, the exchange of the two nucleons will multiply the deuterium wavefunction by (-1) from isospin exchange, (-1) from spin exchange and (-1) from parity (location exchange), again for a total of (-1) as needed for antisymmetry.

In the first case the deuteron has is a Spin triplet, so that its total spin s is 1. It also has an even parity and therefore even orbital angular momentum, the lower its energy. Therefore the lowest possible energy state has s =1, l =0.

In the second case the deuteron has is a spin singlet, so that its total spin s is 0. It also has an odd parity and therefore odd orbital angular momentum l . Therefore the lowest possible energy state has s =0, l =1.

Since s =1 gives a stronger nuclear attraction, the deuterium ground state is in the s =1, l =0 state.

The same considerations lead to the possible states of an isospin triplet having s =0, l =even or s =1, l =odd. Thus the state of lowest energy has s =1, l =1, higher than that of the isospin singlet.

The analysis just given is in fact only approximate, both because isospin is not an exact symmetry, and more importantly because the strong nuclear interaction between the two a way that mixes different s and l states. That is, s and l are not constant in time (they do not commute with the Hamiltonian), and over time a state such as s =1, l =0 may become a state of s =1, l =2. Parity is still constant in time so these do not mix with odd l states (such as s =0, l =1). Therefore the quantum state of the deuterium is a superposition (a linear combination) of the s =1, l =0 state and the s =1, l =2 state, even though the first component is much bigger. Since the total angular momentum j is also a good quantum number (it is a constant in time), both components must have the same j, and therefore j =1. This is the total spin of the deuterium nucleus.

To summarize, the deuterium nucleus is antisymmetric in terms of isospin, and has spin 1 and even (+1) parity. The relative angular momentum of its nucleons l is not well defined, and the deuterium is a superposition of mostly l =0 with some l =2.

Magnetic and electric multipoles

In order to find theoretically the deuterium magnetic dipole moment $μ$ , one uses the formula for a nuclear magnetic moment

$\mu = {1\over (j+1)}\langle(l,s),j,m_j=j|\overrightarrow{\mu}\cdot \overrightarrow{j}|(l,s),j,m_j=j\rangle$

with

$\overrightarrow{\mu} = g^{(l)}\overrightarrow{l} + g^{(s)}\overrightarrow{s}$

g^(l) and g^(s) are nucleons.

Since the orbital angular momentum $\overrightarrow{l}$ and spin $\overrightarrow{s}$ . One arrives at

$\mu = {1\over (j+1)}\langle(l,s),j,m_j=j|\left({1\over 2}\overrightarrow{l} {g^{(l)}}_p + {1\over 2}\overrightarrow{s} ({g^{(s)}}_p + {g^{(s)}}_n)\right)\cdot \overrightarrow{j}|(l,s),j,m_j=j\rangle$

where subscripts p and n stand for the proton and neutron, and g^(l)_n = 0.

By using the same identities as here and using the value g^(l)_p = 1 in nuclear magneton units, we arrive at the following result, in nuclear magneton units

$\mu = {1\over 4 (j+1)}\left[({g^{(s)}}_p + {g^{(s)}}_n)\big(j(j+1) - l(l+1) + s(s+1)\big) + \big(j(j+1) + l(l+1) - s(s+1)\big)\right]$

For the s =1, l =0 state j =1 and we get, in nuclear magneton units

$\mu = {1\over 2}({g^{(s)}}_p + {g^{(s)}}_n) = 0.879$

For the s =1, l =2 state with j =1 we get, in nuclear magneton units

$\mu = -{1\over 4}({g^{(s)}}_p + {g^{(s)}}_n) + {3\over 4} = 0.310$

The measured value of the deuterium magnetic dipole moment, in nuclear magneton units, is 0.857. This suggests that the state of the deuterium is indeed only approximately s =1, l =0 state, and is actually a linear combination of (mostly) this state with s =1, l =2 state.

The electric dipole is zero as usual.

The measured electric quadropole of the deuterium is 0.2859 e fm², where e is the electric charge is e, the above model does not suffice for its computation. More specifically, the electric quadropole does not get a contribution from the l =0 state (which is the dominant one) and does get a contribution from a term mixing the l =0 and the l =2 states, because the electric quadrupole operator does not commute with angular momentum. The latter contribution is dominant in the absence of a pure l =0 contribution, but cannot be calculated without knowing the exact spatial form of the nucleons wavefunction inside the deuterium.

Higher magnetic and electric multipole moments cannot be calculated by the above model, for similar reasons.

Deuterium radius

Further information: Nuclear size

The square root of the average squared radius of the deuterium, measured experimentally, is $\sqrt{\langle r^2 \rangle} = 0.96$ fermi (= 0.96 fm).

Applications

Deuterium is useful in protium, deuterium undergoes fusion purely via the strong interaction, making its use for commercial power plausible.

In infrared spectrometry.

Neutron scattering techniques particularly profit from availability of deuterated samples: The H and D cross sections are very distinct and different in sign, which allows contrast variation in such experiments. Further, a nuisance problem of ordinary hydrogen is its large incoherent neutron cross section, which is nil for D and delivers much clearer signals in deuterated samples. Hydrogen occurs in all materials of organic chemistry and life science, but cannot be seen by X-ray diffraction methods. Hydrogen can be seen by neutron diffraction and scattering, which makes neutron scattering, together with a modern deuteration facility, indispensable for many studies of macromolecules in biology and many other areas.

Deuterium is useful in hydrogen nuclear magnetic resonance spectroscopy (proton NMR). NMR ordinarily requires compounds of interest to be analyzed as dissolved in solution. Because of deuterium's nuclear spin properties which differ from the light hydrogen usually present in organic molecules, NMR spectra of hydrogen/protium are highly differentiable from that of deuterium, and in practice deuterium is not "seen" by an NMR instrument tuned to light-hydrogen. Deuterated solvents (including heavy water, but also compounds like deuterated chloroform CDCl₃) are therefore routinely used in NMR spectroscopy, in order to allow only the light-hydrogen spectra of the compound of interest to be measured, without solvent-signal interference.

Deuterium can also be used for femtosecond infrared spectroscopy, since the mass difference drastically affects the frequency of molecular vibrations; deuterium-carbon bond vibrations are found in locations free of other signals.

Measurements of small variations in the natural abundances of deuterium, along with those of the stable heavy oxygen isotopes ¹⁷O and ¹⁸O, are of importance in global meteoric water line (GMWL). This plot allows samples of precipitation-originated water to be identified along with general information about the climate in which it originated. Evaporative and other processes in bodies of water, and also ground water processes, also differentially alter the ratios of heavy hydrogen and oxygen isotopes in fresh and salt waters, in characteristic and often regionally-distinctive ways.^[5]

The proton and neutron making up deuterium can be dissociated through neutral current interactions with cross section for this interaction is comparatively large, and deuterium was successfully used as a neutrino target in the Sudbury Neutrino Observatory experiment.

History

Lighter element isotopes suspected

The existence of nonradioactive isotopes of lighter elements had been suspected in studies of neon as early as 1913, and proven by mass spectroscopy of light elements in 1920. The prevailing theory at the time, however, was that the isotopes were due to the existence of differing numbers of "nuclear electrons" in different atoms of an element. It was expected that hydrogen, with a measured average atomic mass very close to 1 u, and a nucleus thought to be composed of a single proton (a known particle), could not contain nuclear electrons, and thus could have no heavy isotopes.

Deuterium predicted and finally detected

Deuterium was predicted in 1926 by Walter Russell, using his "spiral" periodic table. It was first detected spectroscopically in late 1931 by chemistry in 1934.

"Heavy water" experiments in World War II

Shortly before the war, Hans von Halban and Lew Kowarski moved their research on neutron moderation from France to England, smuggling the entire global supply of heavy water (made in Norway) across in twenty-six steel drums.^[6]^[7]

During World War II, Nazi Germany was known to be conducting experiments using plutonium for an atomic bomb. Ultimately, it led to (what seemed to be important at that time) the Allied operation called the "Norwegian heavy water sabotage," the purpose of which was to destroy the Vemork deuterium production/enrichment facility in Norway.

After World War II ended, the Allies discovered that Germany was not putting as much serious effort into the program as has had been previously thought. The Germans had completed only a small, partly-built experimental reactor (which had been hidden away). By the end of the war, the Germans did not even have a fifth the amount of heavy water needed to run the reactor, partially due to the Norwegian heavy water sabotage operation. However, even had the Germans succeeded in getting a reactor operational (as the U.S. did with a graphite reactor in late 1942), they would still have been at least several years away from development of an atomic bomb with maximal effort. The engineering process, even with maximal effort and funding, required about two and a half years (from first critical reactor to bomb) in both the U.S. and U.S.S.R, for example (see the article heavy water for a more complete history of its production and use).

Data

Density: 0.180 kg/m³ at STP (0 °C, 101.325 kPa).
Atomic weight: 2.01355321270.
Mean abundance in ocean water (see VSMOW) about 0.0156 % of H atoms = 1/6400 H atoms.

Data at approximately 18 K for D₂ (triple point):

Density:

Liquid: 162.4 kg/m³
Gas: 0.452 kg/m³

Viscosity: 1.3 µPa·s
Specific heat capacity at constant pressure c_p:

Solid: 2950 J/(kg·K)
Gas: 5200 J/(kg·K)

Anti-deuterium

An antideuteron is the antiparticle of the nucleus of deuterium, consisting of an positron orbiting the nucleus, would be called antideuterium, but as of 2005 antideuterium has not yet been created. The symbol for antideuterium is the same as for deuterium, except with a bar over it.

References

Notes

^ "Hubble measures deuterium on Jupiter", Science News – Find Articles, 5 October 1996. Retrieved on 2007-09-10.
^ Deuteros at Studylight.org
^ § IR-3.3.2, Provisional Recommendations, Nomenclature of Inorganic Chemistry, Chemical Nomenclature and Structure Representation Division, IUPAC. Accessed on line October 3, 2007.
^ IUPAC Commission on Nomenclature of Inorganic Chemistry (2001). "Names for Muonium and Hydrogen Atoms and their Ions" (PDF). Pure and Applied Chemistry 73: 377–380.
^ Oxygen – Isotopes and Hydrology. SAHRA. Retrieved on 2007-09-10.
^ Sherriff, Lucy (2007-06-01). Royal Society unearths top secret nuclear research. The Register. Situation Publishing Ltd.. Retrieved on 2007-06-03.
^ The Battle for Heavy Water Three physicists' heroic exploits. CERN Bulletin. European Organization for Nuclear Research (2002-04-01). Retrieved on 2007-06-03.
^ Massam, T., et al. (1965). "Experimental observation of antideuteron production". Il Nuovo Cimento 39: 10–14.
^ Dorfan, D. E., et al. (June 1965). "Observation of Antideuterons". Phys. Rev. Lett. 14 (24): 1003–1006. doi:10.1103/PhysRevLett.14.1003.

General references

Nuclear Data Evaluation Lab
Mullins, Justin (27 April 2005). "Desktop nuclear fusion demonstrated with deuterium gas". New Scientist. Retrieved on 2007-09-10.
Annotated bibliography for Deuterium from the Alsos Digital Library for Nuclear Issues
Missing Gas Found in Milky Way. Space.combe-x-old:Дэўтэр

Categories: Nuclear fusion fuels

This article is licensed under the GNU Free Documentation License. It uses material from the Wikipedia article "Deuterium". A list of authors is available in Wikipedia.