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Electronvolt
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=== Momentum === By dividing a particle's kinetic energy in electronvolts by the fundamental constant ''c'' (the speed of light), one can describe the particle's [[momentum]] in units of eV/''c''.<ref name="FNALunits">{{cite web |url=http://quarknet.fnal.gov/toolkits/ati/whatgevs.html |title=Units in particle physics |publisher=Fermilab |date=22 March 2002 |work=Associate Teacher Institute Toolkit |access-date=13 February 2011 |url-status=live |archive-url=https://web.archive.org/web/20110514152552/http://quarknet.fnal.gov/toolkits/ati/whatgevs.html |archive-date=14 May 2011 }}</ref> In natural units in which the fundamental velocity constant ''c'' is numerically 1, the ''c'' may informally be omitted to express momentum using the unit electronvolt. [[File:Einstein-triangle-in-natural-units.svg|thumb|The [[energy–momentum relation]] in [[natural units]], <math>E^2 = p^2 + m_0^2</math>, is a [[Pythagorean theorem|Pythagorean equation]] that can be visualized as a [[right triangle]] where the total [[energy]] <math>E</math> is the [[hypotenuse]] and the [[momentum]] <math>p</math> and [[Invariant mass|rest mass]] <math>m_0</math> are the two [[Cathetus|legs]].]] The [[energy–momentum relation]] <math display="block">E^2 = p^2 c^2 + m_0^2 c^4</math> in natural units (with <math>c=1</math>) <math display="block">E^2 = p^2 + m_0^2</math> is a [[Pythagorean equation]]. When a relatively high energy is applied to a particle with relatively low [[rest mass]], it can be approximated as <math>E \simeq p</math> in [[Particle physics|high-energy physics]] such that an applied energy with expressed in the unit eV conveniently results in a numerically approximately equivalent change of momentum when expressed with the unit eV/''c''. The dimension of momentum is {{dimanalysis|length=1|mass=1|time=−1}}. The dimension of energy is {{dimanalysis|length=2|mass=1|time=−2}}. Dividing a unit of energy (such as eV) by a fundamental constant (such as the speed of light) that has the dimension of velocity ({{dimanalysis|length=1|time=−1}}) facilitates the required conversion for using a unit of energy to quantify momentum. For example, if the momentum ''p'' of an electron is {{val|1|u=GeV/''c''}}, then the conversion to [[MKS system of units]] can be achieved by: <math display="block">p = 1\; \text{GeV}/c = \frac{(1 \times 10^9) \times (1.602\ 176\ 634 \times 10^{-19} \; \text{C}) \times (1 \; \text{V})}{2.99\ 792\ 458 \times 10^8\; \text{m}/\text{s}} = 5.344\ 286 \times 10^{-19}\; \text{kg} {\cdot} \text{m}/\text{s}.</math>
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