Read the entire row
Quantity and symbol identify physical meaning; a number alone is insufficient. h has units J·s, k connects temperature to energy and G enters Newtonian gravitation. A table supplies values rather than assumptions of every equation. A precise G value, for instance, does not make the point-mass force law valid for arbitrary extended geometry.
CODATA 2022 names the adjustment rather than the date of an individual experiment. Recommended values form a consistent set. If a problem explicitly supplies a rounded value, use and identify that assumption; avoid combining older and newer high-precision values without justification.[3]
What is exact since 2019
The SI fixes seven defining constants: ΔνCs, c, h, e, k, Na and Kcd. Their assigned numerical values have no uncertainty within the definition. Real time intervals, voltages and temperatures are nevertheless realised and measured with finite uncertainty. Exact unit anchors do not make instruments or input data perfect.[2]
The defining Kcd value applies at frequency 540×1012 Hz. The value 683 lm/W is not the universal efficacy of every lamp. Likewise, c denotes vacuum light speed; propagation in matter requires an appropriate optical model. Definition status does not establish that a constant applies to a particular situation.[2]
Exact derived constants
Products or ratios of exact factors remain exact. R = 8.31446261815324 J·mol−1·K−1 and F = 96485.3321233100184 C·mol−1 have finite exact decimal values here. The decimal expansion of ℏ does not terminate: its displayed value is rounded while its defining relationship is exact.[1]
Exact therefore does not mean every printed digit string is the complete number. An approximation sign can label a shortened representation of an exactly determined quantity. If many digits are needed, calculate h/(2π) with adequate precision rather than treating a short ℏ display as an independent measurement.
Measured values and parenthetical notation
The recommended G value is 6.67430(15)×10−11 m3·kg−1·s−2. Parentheses refer to the final digits: u(G)=0.00015×10−11 = 1.5×10−15 m3·kg−1·s−2. This is a standard uncertainty rather than a strict maximum error. Its relative value u(G)/G is approximately 2.25×10−5, or 22.5 parts per million.[1]
Particle masses in kilograms, the atomic mass constant and the fine-structure constant α remain experimentally determined. μ0 and ε0 are also no longer exact in the present SI; they depend on measured α among other quantities. The former exact assignment μ0=4π×10−7 N/A2 must not be presented as a current exact value.[3]
Powers, units and signs
Keep the power of ten attached to the number. 6.626×10−34 is not 6.626×1034: reversing this exponent changes the result by 68 orders of magnitude. In E=hf, J·s times s−1 gives J. Elementary charge e is positive; an electron has q=−e. A positive constant in the table does not determine the charge sign.
An electronvolt is an energy unit rather than a voltage unit. Its conversion factor is exact; a measured particle energy may still carry uncertainty.[1]
Example: photon energy with measurement uncertainty
Energy and frequency uncertainty
- Given
- f=(6.40±0.02)×1014 Hz, with ± denoting standard uncertainty. Use exact h.
- Find
- Find E in J and eV, including u(E).
Apply E=hf; h adds no uncertainty.
Calculate E=4.240684896×10−19 J and divide by e to obtain eV.
Because the model is linear, u(E)=h u(f)=1.32521403×10−21 J =0.0082713354 eV.
Result: A suitable report is E=(4.241±0.013)×10−19 J or (2.647±0.008) eV.
Check: Back-substitution E/h gives 6.40×1014 Hz. Both relative uncertainties are 0.3125%. Rounding is not a new measurement uncertainty.
Retained introductory example
- Given
- f=5.00×1014 Hz; h exact.
- Find
- Find E.
Apply E=hf.
J·s times s−1 gives J.
Result: E=3.313035075×10−19 J, or 3.31×10−19 J to three significant figures.
Check: Dividing unrounded E by h returns the specified f. No experimental u(E) can be calculated without an input uncertainty.
Example: deriving R and using an ideal gas
Pressure from amount and temperature
- Given
- Model inputs n=0.0800 mol, T=310 K, V=2.00 L. Assume an ideal gas and initially treat inputs as specified.
- Find
- Derive R and find absolute pressure.
Nak gives 8.31446261815324 J/(mol·K).
Convert 2.00 L to 0.00200 m3.
The product nRT=206.19867293 J; dividing by V gives 103099.3365 Pa.
Result: The model gives approximately 103 kPa absolute; exact R does not remove limitations of ideal-gas assumptions or real inputs.
Check: pV=206.19867293 Pa·m3=206.19867293 J agrees with nRT. Doubling V would halve p.
Propagate uncertainty and account for correlation
For y=f(x), this is the linear approximation at the working point. The covariance term disappears only for uncorrelated input estimates. A CODATA adjustment supplies interconnected constants; inspect their correlations when combining them precisely.[4]
If G is the only uncertain factor in F=Gm1m2/r2, its relative uncertainty transfers directly to F when other quantities are fixed. Measured masses and distance also contribute; the squared distance gives sensitivity factor 2. An expanded interval U=ku additionally requires a stated coverage factor.[4][5]
Document precision and model limitations
Extra digits can be useful internally but add no experimental information. Rounding at the end avoids unnecessary loss of precision. Report standard uncertainty with value, unit and data edition. “Exact” describes a definition; measurement accuracy concerns the actual observation. A fixed constant used in an unsuitable equation can still produce an incorrect result.
Keep applicability explicit: c refers to vacuum, G to the chosen gravitational theory and h to quantum relationships. Standard uncertainty is not automatically a 95% interval; even the usual interpretation of a single-u interval requires distributional assumptions. Assess model discrepancies separately instead of hiding them behind extra decimal places.[5]
Three problems with solutions
Convert 3.20 eV to joules.
Show hint
Use the exact eV factor.
Show solution and calculation
3.20×1.602176634×10−19 =5.1269652288×10−19 J, or 5.13×10−19 J to three significant figures.
Find uᵣ(G) from 6.67430(15)×10−11.
Show hint
The exponent cancels in the ratio.
Show solution and calculation
0.00015/6.67430≈2.247×10−5=0.002247%=22.47 ppm.
This is standard uncertainty rather than a guaranteed error bound.
Which remains exact: R=Nak or Mᵤ=Namᵤ? Explain in the present SI.
Show hint
Inspect the status of each factor.
Show solution and calculation
R is exact because Na and k are fixed.
Mᵤ is measured because mᵤ in kilograms is measured.
CODATA 2022 recommends Mᵤ=1.00000000105(31)×10−3 kg/mol; 10−3 kg/mol is no longer its exact SI assignment.
Sources for this explanation
Quanta · Updated October 1, 2026
- NIST: Complete CODATA 2022 recommended values
- BIPM: SI Brochure, 9th edition, current 2026 version
- CODATA 2022 adjustment: Mohr et al. (2025)
- NIST: Combining uncertainty components
- NIST: Expanded uncertainty and coverage factors
Editorial responsibility: Quanta Study editorial team
Review scope
This explanation, 3 worked examples and 3 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.
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An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.
Editorial responsibility, sources and methodsCite this reference page
Quanta. Selected fundamental constants. Updated October 1, 2026. https://quanta-study.de/en/formula-sheet/tables/fundamentale-konstanten
When citing a specific scientific statement, also cite the specialist source linked to that statement.