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?
--
73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.co.uk/g3sek
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