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Equation 5 · Part 3 · The Last Artefact: Redefining the Kilogram

Symbol a^3

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)
a3a^{3}

What this part means

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

Its job in the formula

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

The passage around this formula

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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Learn the underlying idea

An exponent tells how a base is used in multiplication. In x³, x is the base and 3 is the exponent: x³ = x × x × x.

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Sources cited in the surrounding passage

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