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Published equation contexts

N=8 Vsa3,ms=N m(28Si)=h N(m(28Si)h)N = \frac{8\,V_{\mathrm s}}{a^{3}}, \qquad m_{\mathrm s} = N\, m({}^{28}\mathrm{Si}) = h\,N \left( \frac{m({}^{28}\mathrm{Si})}{h} \right)

Why this formula appears here

The second primary method shares almost nothing with the first except its answer. The X-ray crystal density route determines the mass of a nearly perfect single-crystal silicon sphere by counting the atoms in it. With eight atoms per cubic unit cell of lattice parameter a, the atom count follows from the macroscopic sphere volume, and the sphere mass follows from the count [ 4 ] : N=8 Vsa3,ms=N m(28Si)=h N(m(28Si)h)N = \frac{8\,V_{\mathrm s}}{a^{3}}, \qquad m_{\mathrm s} = N\, m({}^{28}\mathrm{Si}) = h\,N \left( \frac{m({}^{28}\mathrm{Si})}{h} \right). The ratio of the silicon-28 atomic mass to h is a constant of nature known to high accuracy, so with h fixed the sphere becomes a primary mass standard [ 4 ] . The mise en pratique notes that the second equality is not exact — the total binding energy of the crystal reduces the right-hand side…

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VsV_{\mathrm s}

Symbol V_mathrm s

VmV_mathrm s occurs above the fraction bar. The numerator is divided by the entire denominator below it.

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a3a^{3}

Symbol a^3

a3a^3 occurs below the fraction bar. The quantity above the bar is divided by this expression; zero is excluded as a denominator.

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msm_{\mathrm s}

Symbol m_mathrm s

mmm_mathrm s is an input to the expression that computes the quantity on the left.

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How to interpret it

With a fixed numerator, increasing a nonzero denominator reduces the fraction. Read it with the definitions, units, and assumptions supplied by the article.

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Published contexts (1)

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N=8 Vsa3,ms=N m(28Si)=h N(m(28Si)h)N = \frac{8\,V_{\mathrm s}}{a^{3}}, \qquad m_{\mathrm s} = N\, m({}^{28}\mathrm{Si}) = h\,N \left( \frac{m({}^{28}\mathrm{Si})}{h} \right)

Equation 5 · Metrology

The Last Artefact: Redefining the Kilogram

This equation states an equality: the expressions on both sides have the same value under the article’s assumptions.

The second primary method shares almost nothing with the first except its answer. The X-ray crystal density route determines the mass of a nearly perfect single-crystal silicon sphere by counting the atoms in it. With eight atoms per cubic unit cell of lattice parameter a, the atom count follows from the macroscopic sphere volume, and the sphere mass follows from the count [ 4 ] : N=8 Vsa3,ms=N m(28Si)=h N(m(28Si)h)N = \frac{8\,V_{\mathrm s}}{a^{3}}, \qquad m_{\mathrm s} = N\, m({}^{28}\mathrm{Si}) = h\,N \left( \frac{m({}^{28}\mathrm{Si})}{h} \right). The ratio of the silicon-28 atomic mass to h is a constant of nature known to high accuracy, so with h fixed the sphere becomes a primary mass standard [ 4 ] . The mise en pratique notes that the second equality is not exact — the total binding energy of the crystal reduces the right-hand side…

Meanings in this article

  • hh: a constant of nature known to high accuracy, so with h fixed the sphere becomes a primary mass standard [ 4 ].
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