What is the Rate of Tungsten Evaporation in a Light Bulb?

AI Thread Summary
Tungsten evaporation in a light bulb occurs as billions of electrons flow through the filament, causing it to heat and emit light. The discussion focuses on calculating the energy required to evaporate a small volume of tungsten and understanding how the current changes as the filament evaporates. It is suggested that a constant fraction of electric power is used for breaking atomic bonds, while the rest is for heating. The thickness of the tungsten wire is not uniform and must be modeled as a function of position and time, with initial conditions considered. The conversation emphasizes the need to relate bond density to surface area due to evaporation occurring from the surface.
tdog123
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Homework Statement


Ina light bulb, billions of electrons flow through the tungsten wire bumping around, causing the filament to glow hot. Hot tungsten evaporates slowly, so its initial thickness profile r(x, t=0) can change over time. Consider a volume V of tungsten, with N bonds total, each requiring energy E to break. We will (rather incorrectly) assume that a constant fraction f of electric power is spent on breaking these bonds, and the remaining (1-f) is spent to heat and illuminate. We will also assume that the temperature remains constant.

a. How much energy must be spent to evaporate a volume dV?
b. As the wire evaporates over time, does the current increase or decrease? Is there anything in this system that remains constant?
c. Derive an equation governing the thickness r(x, t) of the tungsten wire as a function of position and time. Don't assume that the thickness is uniform.
d. Assume now that the thickness r(x, t) has no x dependence (i.e. we start with a perfectly cylindrical wire). Solver for r(t).
e. plot, and discuss your results

notes:
Since the evaporation happens from the surface, you might find it helpful to think in terms of n = number of bonds per area, instead of number of bonds per volume. It's perfectly ok if you write your answer in terms of n, instead of number of bonds per volume.

Homework Equations


Not that I know of

The Attempt at a Solution


Don't even know how to start with this problem
 
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tdog123 said:
Don't even know how to start with this problem
Take it one step at a time. a. is quite easy. If a volume V has N bonds, how many does a volume dV have? How much energy to break them all?
 
dN? so I need to create a function which relates N to dt? Okay so would that look something like fPt? or do I need to relate it to surface area because it evaporates from the outside in?
 
tdog123 said:
dN? so I need to create a function which relates N to dt? Okay so would that look something like fPt? or do I need to relate it to surface area because it evaporates from the outside in?
No, the answer to the first of my two questions only involves the three variables provided, V, N, and dV.
 
In a volume dV there are (N/V)dV bonds. So it takes (N/V)dV*E energy to break them
 
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What about the thickness equation though? I think that's the hardest part.
 
limeset said:
What about the thickness equation though? I think that's the hardest part.
Sure, but you need an answer to b first.
 
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