Editing Deuterium (section)

== Properties ==

=== Data for molecular deuterium ===
Formula: {{chem2|D2}} or {{chem|2|1|H|2}}

* Density: 0.180 kg/m{{sup|3}} at [[Standard temperature and pressure|STP]] (0 °C, 101325 Pa).
* Atomic weight: 2.0141017926 Da.
* Mean abundance in ocean water (from [[Vienna Standard Mean Ocean Water|VSMOW]]) 155.76 ± 0.1 atoms of deuterium per million atoms of all isotopes of hydrogen (about 1 atom of in 6420); that is, about 0.015% of all atoms of hydrogen (any isotope)

Data at about 18 K for {{sup|2}}H{{sub|2}} ([[triple point]]):
* Density:
** Liquid: 162.4 kg/m{{sup|3}}
** Gas: 0.452 kg/m{{sup|3}}
** Liquefied {{sup|2}}H{{sub|2}}O: 1105.2 kg/m{{sup|3}} at [[Standard temperature and pressure|STP]]
* Viscosity: 12.6 [[pascal second|μPa·s]] at 300 K (gas phase)
* Specific heat capacity at constant pressure ''c{{sub|p}}'':
** Solid: 2950 J/(kg·K)
** Gas: 5200 J/(kg·K)

=== Physical properties ===

Compared to hydrogen in its natural composition on Earth, pure deuterium ({{sup|2}}H{{sub|2}}) has a higher [[melting point]] (18.72&nbsp;K vs. 13.99&nbsp;K), a higher [[boiling point]] (23.64 vs. 20.27&nbsp;K), a higher [[Critical point (thermodynamics)|critical temperature]] (38.3 vs. 32.94&nbsp;K) and a higher critical pressure (1.6496 vs. 1.2858&nbsp;MPa).<ref>{{cite web |title=Deuterium, {{sup|2}}H |series=compounds |website=PubChem |publisher=U.S. [[National Institutes of Health]] |url=https://pubchem.ncbi.nlm.nih.gov/compound/deuterium}}</ref>

The physical properties of deuterium compounds can exhibit significant [[kinetic isotope effect]]s and other physical and chemical property differences from the protium analogs. {{sup|2}}H{{sub|2}}O, for example, is more [[viscous]] than normal {{H2O|link=y}}.<ref>{{RubberBible86th}}</ref> There are differences in bond energy and length for compounds of heavy hydrogen isotopes compared to protium, which are larger than the isotopic differences in any other element. Bonds involving deuterium and [[tritium]] are somewhat stronger than the corresponding bonds in protium, and these differences are enough to cause significant changes in biological reactions. Pharmaceutical firms are interested in the fact that {{sup|2}}H is harder to remove from carbon than {{sup|1}}H.<ref>{{cite news |last=Halford |first=Bethany | name-list-style = vanc |date=4 July 2016 |title=The deuterium switcheroo |newspaper=[[Chemical & Engineering News]] |volume=94 |issue=27 |publisher=[[American Chemical Society]] |pages=32–36 | doi = 10.1021/cen-09427-cover}}</ref>

Deuterium can replace {{sup|1}}H in water molecules to form heavy water ({{sup|2}}H{{sub|2}}O), which is about 10.6% denser than normal water (so that ice made from it sinks in normal water). Heavy water is slightly toxic in [[eukaryotic]] animals, with 25% substitution of the body water causing cell division problems and sterility, and 50% substitution causing death by cytotoxic syndrome (bone marrow failure and gastrointestinal lining failure). [[Prokaryotic]] organisms, however, can survive and grow in pure heavy water, though they develop slowly.<ref>{{cite journal | vauthors = Kushner DJ, Baker A, Dunstall TG | date = February 1999 | title = Pharmacological uses and perspectives of heavy water and deuterated compounds | journal = Canadian Journal of Physiology and Pharmacology | volume = 77 | issue = 2 | pages = 79–88 | pmid = 10535697 | doi = 10.1139/cjpp-77-2-79 }}</ref> Despite this toxicity, consumption of heavy water under normal circumstances does not pose a [[heavy water#Toxicity in humans|health threat]] to humans. It is estimated that a {{convert|70|kg|lb|0|abbr=on}} person might drink {{convert|4.8|L|USgal|abbr=out}} of heavy water without serious consequences.<ref>{{Cite book |chapter-url=https://books.google.com/books?id=nQh7iGX1geIC&pg=PA111 |isbn=978-1-4020-1314-0 |pages=111–112 |chapter=Physiological effect of heavy water |editor=Vertes, Attila |year=2003 |publisher=Kluwer |location=Dordrecht |title=Elements and isotopes: formation, transformation, distribution}}</ref> 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 [[quantum spin|spin]] +1 ("[[triplet state]]") and is thus a [[boson]]. The [[nuclear magnetic resonance|NMR]] frequency of deuterium is significantly different from normal hydrogen. [[Infrared spectroscopy]] also easily differentiates many deuterated compounds, due to the large difference in IR absorption frequency seen in the vibration of a chemical bond containing deuterium, versus light hydrogen. The two stable isotopes of hydrogen can also be distinguished by using [[mass spectrometry]].

The triplet deuteron nucleon is barely bound at {{nowrap|1=''E''{{sub|B}} = {{val|2.23|u=MeV}}}}, and none of the higher energy states are bound. The singlet deuteron is a virtual state, with a negative binding energy of {{val|p=~|60|u=keV}}. There is no such stable particle, but this virtual particle transiently exists during neutron–proton inelastic scattering, accounting for the unusually large neutron scattering cross-section of the proton.<ref>{{cite web |title=Neutron–proton scattering |date=Fall 2004 |website=mightylib.mit.edu |type=course notes |id=22.101 |publisher=[[Massachusetts Institute of Technology]] |url=http://mightylib.mit.edu/Course%20Materials/22.101/Fall%202004/Notes/Part3.pdf |url-status=dead <!-- presumed --> |access-date=2011-11-23 |archive-url=https://web.archive.org/web/20110721213924/http://mightylib.mit.edu/Course%20Materials/22.101/Fall%202004/Notes/Part3.pdf |archive-date=2011-07-21}}</ref>

=== Nuclear properties (deuteron) ===

==== Deuteron mass and radius ====
The deuterium nucleus is called a '''deuteron'''. It has a mass of {{physconst|md_Da}} (just over {{val|1.875|u=GeV/c2}}<ref>{{cite web |title=Deuteron mass energy equivalent in MeV |website=Physics.nist.gov |publisher=U.S. [[National Institute of Standards and Technology]] |url=https://physics.nist.gov/cgi-bin/cuu/Value?mdc2mev |access-date=2024-05-30}}</ref>).

The [[charge radius]] of a deuteron is {{physconst|rd|after=.}}

Like the [[Proton#Charge radius|proton radius]], measurements using [[muon]]ic deuterium produce a smaller result: {{val|2.12562|(78)|ul=fm}}.<ref>{{cite journal | vauthors = Pohl R, Nez F, Fernandes LM, Amaro FD, Biraben F, Cardoso JM, Covita DS, Dax A, Dhawan S, Diepold M, Giesen A, Gouvea AL, Graf T, Hänsch TW, Indelicato P, Julien L, Knowles P, Kottmann F, Le Bigot EO, Liu YW, Lopes JA, Ludhova L, Monteiro CM, Mulhauser F, Nebel T, Rabinowitz P, dos Santos JM, Schaller LA, Schuhmann K, Schwob C, Taqqu D, Veloso JF, Antognini A | display-authors = 6 | date = August 2016 | title = Laser spectroscopy of muonic deuterium | journal = [[Science (journal)|Science]] | volume = 353 | issue = 6300 | pages = 669–673 | pmid = 27516595 | doi = 10.1126/science.aaf2468 | hdl-access = free | collaboration = The CREMA Collaboration | bibcode = 2016Sci...353..669P | hdl = 10316/80061 | s2cid = 206647315 }}</ref>

{{see also|Proton radius puzzle}}

==== Spin and energy ====

Deuterium is one of only five stable [[nuclide]]s with an odd number of protons and an odd number of neutrons. ({{sup|2}}H, [[isotopes of lithium|{{sup|6}}Li]], [[isotopes of boron|{{sup|10}}B]], [[isotopes of nitrogen|{{sup|14}}N]], [[isotopes of tantalum|{{sup|180m}}Ta]]; the long-lived radionuclides [[Potassium-40|{{sup|40}}K]], [[isotopes of vanadium|{{sup|50}}V]], [[isotopes of lanthanum|{{sup|138}}La]], [[isotopes of lutetium|{{sup|176}}Lu]] also occur naturally.) Most [[odd–odd nuclei]] are unstable to [[beta decay]], because the decay products are [[even-even nuclei|even–even]], and thus more strongly bound, due to [[Semi-empirical mass formula#Pairing term|nuclear pairing effects]]. Deuterium, however, benefits from having its proton and neutron coupled to a spin-1 state, which gives a stronger nuclear attraction; the corresponding spin-1 state does not exist in the two-neutron or two-proton system, due to the [[Pauli exclusion principle]] which would require one or the other identical particle with the same spin to have some other different quantum number, such as [[angular momentum operator|orbital angular momentum]]. But orbital angular momentum of either particle gives a lower [[binding energy]] for the system, mainly due to increasing distance of the particles in the steep gradient of the nuclear force. In both cases, this causes the [[diproton]] and [[dineutron]] to be [[unstable]].

The proton and neutron in deuterium can be [[Dissociation (chemistry)|dissociated]] through [[neutral current]] interactions with [[neutrino]]s. The [[Cross section (physics)|cross section]] for this interaction is comparatively large, and deuterium was successfully used as a neutrino target in the [[Sudbury Neutrino Observatory]] experiment.

Diatomic deuterium ({{sup|2}}H{{sub|2}}) has ortho and para [[Spin isomers of hydrogen|nuclear spin isomers]] like diatomic hydrogen, but with [[Spin isomers of hydrogen#Deuterium|differences in the number and population of spin states and rotational levels]], which occur because the deuteron is a [[boson]] with nuclear spin equal to one.<ref name=Hollas>{{cite book |last1=Hollas |first1=J. Michael | name-list-style = vanc |title=Modern Spectroscopy |date=1996 |publisher=John Wiley and Sons |isbn=0-471-96523-5 |page=115 |edition=3rd}}</ref>

====Isospin singlet state of the deuteron ====
Due to the similarity in mass and nuclear properties between the proton and neutron, they are sometimes considered as two symmetric types of the same object, a [[nucleon]]. While only the proton has electric charge, this is often negligible due to the weakness of the [[electromagnetic interaction]] relative to the [[strong nuclear interaction]]. The symmetry relating the proton and neutron is known as [[isospin]] and denoted ''I'' (or sometimes ''T'').

Isospin is an [[SU(2)]] symmetry, like ordinary [[Spin (physics)|spin]], so is completely analogous to it. The proton and neutron, each of which have iso[[spin-1/2]], form an isospin doublet (analogous to a [[spin doublet]]), with a "down" state (↓) being a neutron and an "up" state (↑) being a proton.{{Citation needed|date=November 2019}} A pair of nucleons can either be in an antisymmetric state of isospin called [[Singlet state|singlet]], or in a symmetric state called [[Spin triplet|triplet]]. In terms of the "down" state and "up" state, the singlet is
: <math>\frac{1}{\sqrt{2}}\Big( |{\uparrow\downarrow}\rangle - |{\downarrow\uparrow}\rangle\Big).</math>, which can also be written :<math>\frac{1}{\sqrt{2}}\Big( |p n \rangle - |n p \rangle\Big).</math>

This is a nucleus with one proton and one neutron, i.e. a deuterium nucleus. The triplet is
: <math>
\left(
\begin{array}{ll}
|{\uparrow\uparrow}\rangle\\
\frac{1}{\sqrt{2}}( |{\uparrow\downarrow}\rangle + |{\downarrow\uparrow}\rangle )\\
|{\downarrow\downarrow}\rangle
\end{array}
\right)
</math>
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 protons, and a nucleus with two neutrons. These states are not stable.

==== Approximated wavefunction of the deuteron ====
The deuteron wavefunction must be antisymmetric if the isospin representation is used (since a proton and a neutron are not identical particles, the wavefunction need not be antisymmetric in general). Apart from their isospin, the two nucleons also have spin and spatial distributions of their wavefunction. The latter is symmetric if the deuteron is symmetric under [[Parity (physics)|parity]] (i.e. has an "even" or "positive" parity), and antisymmetric if the deuteron is antisymmetric under parity (i.e. has an "odd" or "negative" parity). The parity is fully determined by the total orbital angular momentum of the two 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 is a spin triplet, so that its total spin ''s'' is 1. It also has an even parity and therefore even orbital angular momentum ''l''. The lower its orbital angular momentum, the lower its energy. Therefore, the lowest possible energy state has {{nowrap|''s'' {{=}} 1}}, {{nowrap|''l'' {{=}} 0}}.

In the second case the deuteron 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 {{nowrap|''s'' {{=}} 0}}, {{nowrap|''l'' {{=}} 1}}.

Since {{nowrap|''s'' {{=}} 1}} gives a stronger nuclear attraction, the deuterium [[ground state]] is in the {{nowrap|''s'' {{=}} 1}}, {{nowrap|''l'' {{=}} 0}} state.

The same considerations lead to the possible states of an isospin triplet having {{nowrap|''s'' {{=}} 0}}, {{nowrap|''l'' {{=}} even}} or {{nowrap|''s'' {{=}} 1}}, {{nowrap|''l'' {{=}} odd}}. Thus, the state of lowest energy has {{nowrap|''s'' {{=}} 1}}, {{nowrap|''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 nucleons is related to [[angular momentum]] in [[spin–orbit interaction]] that mixes different ''s'' and ''l'' states. That is, ''s'' and ''l'' are not constant in time (they do not [[commutativity|commute]] with the [[Hamiltonian (quantum mechanics)|Hamiltonian]]), and over time a state such as {{nowrap|''s'' {{=}} 1}}, {{nowrap|''l'' {{=}} 0}} may become a state of {{nowrap|''s'' {{=}} 1}}, {{nowrap|''l'' {{=}} 2}}. Parity is still constant in time, so these do not mix with odd ''l'' states (such as {{nowrap|''s'' {{=}} 0}}, {{nowrap|''l'' {{=}} 1}}). Therefore, the [[quantum state]] of the deuterium is a [[Quantum superposition|superposition]] (a linear combination) of the {{nowrap|''s'' {{=}} 1}}, {{nowrap|''l'' {{=}} 0}} state and the {{nowrap|''s'' {{=}} 1}}, {{nowrap|''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 {{nowrap|''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 deuteron is a superposition of mostly {{nowrap|''l'' {{=}} 0}} with some {{nowrap|''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]]
: <math>\mu = 
\frac{1}{j+1}\bigl\langle(l,s),j,m_j{=}j \,\bigr|\, \vec{\mu}\cdot \vec{\jmath} \,\bigl|\,(l,s),j,m_j{=}j\bigr\rangle</math>
with
: <math>\vec{\mu} = g^{(l)}\vec{l} + g^{(s)}\vec{s} </math>
''g''{{sup|(''l'')}} and ''g''{{sup|(''s'')}} are [[g-factor (physics)|''g''-factor]]s of the nucleons.

Since the proton and neutron have different values for ''g''{{sup|(''l'')}} and ''g''{{sup|(''s'')}}, one must separate their contributions. Each gets half of the deuterium orbital angular momentum <math>\vec{l}</math> and spin <math>\vec{s}</math>. One arrives at
: <math>\mu = 
\frac{1}{j+1} \Bigl\langle(l,s),j,m_j{=}j \,\Bigr|\left(\frac{1}{2}\vec{l} {g^{(l)}}_p + \frac{1}{2}\vec{s} ({g^{(s)}}_p + {g^{(s)}}_n)\right)\cdot \vec{\jmath} \,\Bigl|\, (l,s),j,m_j{=}j \Bigr\rangle</math>
where subscripts p and n stand for the proton and neutron, and {{nowrap|''g''<sup>(l)</sup><sub>n</sub> {{=}} 0}}.

By using the same identities as [[Nuclear magnetic moment#Calculating the magnetic moment|here]] and using the value {{nowrap|g<sup>(l)</sup><sub>p</sub> {{=}} 1}}, one gets the following result, in units of the [[nuclear magneton]] ''μ''{{sub|N}}
: <math>\mu = 
\frac{1}{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]</math>

For the {{nowrap|''s'' {{=}} 1}}, {{nowrap|''l'' {{=}} 0}} state ({{nowrap|''j'' {{=}} 1}}), we obtain
: <math>\mu = \frac{1}{2}({g^{(s)}}_p + {g^{(s)}}_n) = 0.879</math>

For the {{nowrap|''s'' {{=}} 1}}, {{nowrap|''l'' {{=}} 2}} state ({{nowrap|''j'' {{=}} 1}}), we obtain
: <math>\mu = -\frac{1}{4}({g^{(s)}}_p + {g^{(s)}}_n) + \frac{3}{4} = 0.310</math>

The measured value of the deuterium [[magnetic dipole moment]], is {{val|0.857|u=''μ''<sub>N</sub>}}, which is 97.5% of the {{val|0.879|u=''μ''<sub>N</sub>}} value obtained by simply adding moments of the proton and neutron. This suggests that the state of the deuterium is indeed to a good approximation {{nowrap|''s'' {{=}} 1}}, {{nowrap|''l'' {{=}} 0}} state, which occurs with both nucleons spinning in the same direction, but their magnetic moments subtracting because of the neutron's negative moment.

But the slightly lower experimental number than that which results from simple addition of proton and (negative) neutron moments shows that deuterium is actually a linear combination of mostly {{nowrap|''s'' {{=}} 1}}, {{nowrap|''l'' {{=}} 0}} state with a slight admixture of {{nowrap|''s'' {{=}} 1}}, {{nowrap|''l'' {{=}} 2}} state.

The [[electric dipole]] is zero [[Nuclear shell model|as usual]].

The measured electric [[quadrupole]] of the deuterium is {{val|0.2859|u=[[elementary charge|e]]·[[femtometer|fm]]{{sup|2}}}}. While the order of magnitude is reasonable, since the deuteron radius is of order of 1 femtometer (see below) and its [[electric charge]] is e, the above model does not suffice for its computation. More specifically, the [[electric quadrupole]] does not get a contribution from the {{nowrap|1=''l'' = 0}} state (which is the dominant one) and does get a contribution from a term mixing the {{nowrap|1=''l'' = 0}} and the {{nowrap|1=''l'' = 2}} states, because the electric quadrupole [[operator (quantum mechanics)|operator]] does not [[Commutative property#Non-commuting operators in quantum mechanics|commute]] with [[angular momentum]].

The latter contribution is dominant in the absence of a pure {{nowrap|''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 moment]]s cannot be calculated by the above model, for similar reasons.