On Apr 8, 2005, at 12:37 AM, Ian White G3SEK wrote:
> Steve Thompson wrote:
>> R.Measures wrote:
>>> The boiling point of anything in a vacuum is lower, but I do not know
>>> how much lower it is for gold. In any case, it would still be above
>>> the melting point of 1063ºC.
>> Can gold leave the grid structure if it's molten but not boiling?
>>
> Basically, yes - just like water (molten ice) will eventually evaporate
> completely away at room temperature.
>
> Everything has a vapour pressure, which increases with temperature.
> However, the rate of evaporation also increases as the surrounding
> pressure decreases. Materials will evaporate more rapidly in the low
> pressure of a "vacuum" tube, and also the effects of the vapour will be
> more noticeable.
>
> Note that "vacuum" is very much a relative term. Depending on the size
> of the tube and the quality of the vacuum, the "empty" space inside a
> typical tube contains somewhere between a million and a billion gas
> molecules.
>
> Being chemically inert, gold is apparently a very good surface material
> from the viewpoint of electron physics. But it does have the problems
> of
> a low melting point and relatively high vapour pressure, so gold-plated
> grids are not tolerant of overheating.
>
> Now here's another question: if individual gold atoms are simply
> evaporating into the "vacuum" space as a gas, how do they get together
> again to form the celebrated balls?
(
By leaving the grid at a high enough rate to be attracted to each other
by universal gravitation and form into spheres before they
radiation-cool below 1063ºC and solidify. To me, the most notable
thing about gold melt-balls is their size uniformity and the wondrous,
lustrous dusting they impart wherever they alight.
>
>
> --
> 73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB)
> http://www.ifwtech.co.uk/g3sek
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>
Rich Measures, 805.386.3734, AG6K, www.somis.org
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