The synthesis of $^{254}\text{No}$

How is $^{254}\text{No}$ synthesised?

Could you explain the reaction where it is preceded by $^{208}\text{Pb}(^{48}\text{Ca}, 2\text{n})$?

References to articles are well enough—I was somehow unable to find anything sufficiently detailed and informative.

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The notation X(Y,Z)W is a compact way of describing nuclear and particle experiments.

• Particles that appear to the left of the comma (,) are in the initial state and those that appear to the right are in the final state.
• The energy and/or momenta of particles that appear inside the parenthesis are measured. Particles that appear outside have unmeasured energy and/or momentum.

One caveat here: some (or all) unmeasured initial energy and momentum may be deduced on the basis that they represent a fixed target material.

Unobserved final-state particles are often omitted, and sometimes a notation like $X$ is used to imply many possible final state (i.e. an inclusive measurement).

So, when I say that my dissertation looked at $A(e,e'p)$, I mean that I fired an electron beam at fixed nuclear targets (we used $^1\mathrm{H}$ for calibration and acceptance; $^2\mathrm{H}$; $^{12}\mathrm{C}$; and $^{56}\mathrm{Fe}$) and measured the coincident electrons and protons emerging from quasi-elastic scattering events. The recoiling nucleus was unobserved, and other events were cut during analysis.

Similarly the notation above suggests that calcium nuclei were accelerated into a lead target, and the fast ejecta was observed. Those events with exactly two ejected neutrons were selected, leaving a unobserved heavy nucleus assumed to be $^{254}\mathrm{No}$ (the assumption is good if you really understand the measured ejecta).

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Excellent, thank you very much for your answer. –  Harold Cavendish Nov 12 '12 at 20:44