Solving the Temperature of a Mercury Vapour Lamp

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Homework Help Overview

The discussion revolves around determining the temperature at which a mercury vapour lamp emits ultraviolet light at a specific wavelength of 2586 Angstroms. The original poster attempts to apply Wien's displacement law to find this temperature but encounters a discrepancy between their calculated result and the answer provided in their textbook.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the application of Wien's displacement law and question the assumptions made regarding the energy transfer in inelastic collisions between molecules. There are suggestions to consider the Maxwell-Boltzmann distribution for molecular speed and to clarify the appropriate formulas for energy calculations.

Discussion Status

The conversation is ongoing, with participants exploring different interpretations of the problem and discussing relevant physical principles. Some guidance has been offered regarding the use of distributions and laws relevant to the thermal spectrum of photons.

Contextual Notes

There is a noted discrepancy between the original poster's calculation and the textbook answer, which raises questions about the assumptions and methods used in the problem-solving process. Additionally, the context of the discussion includes the application of the inverse square law of radiation in a separate experiment involving point sources.

Amith2006
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Sir,
The question is as follows:
If excitation energy is provided by inelastic collisions between molecules, what is the temperature at which mercury vapour lamp will emit Ultra violet light of wavelength 2586 Angstrom?
I tried applying Wein's displacement law(lamda * T = 2.898 *10^(-3)metre-Kelvin . I got the answer as 11000 Kelvin. But the answer given in my book is 55000 Kelvin.Here the symbol ^ represents power.Please help me in solving this problem.
 
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your answer should come out in kelvin only... not metre K.
For the molecules in the gas they are MOST likely to have some speed (a formula exists for this). When TWO molecules (at a minimum) hit into each other (let's assume that) all their kinetic energy is converted into a photon (whose energy you can calculate).
can you write the appropriate formulae for this?
 
Remember that the speed distribution of the molecules is given by the Maxwell-Boltzmann distribution, not the Planck distribution. Wein's displacement law gives you the peak intensity of a thermal spectrum of photons.
 
Amith2006 said:
Sir,
The question is as follows:
If excitation energy is provided by inelastic collisions between molecules, what is the temperature at which mercury vapour lamp will emit Ultra violet light of wavelength 2586 Angstrom?
I tried applying Wein's displacement law(lamda * T = 2.898 *10^(-3)metre-Kelvin . I got the answer as 11000 Kelvin. But the answer given in my book is 55000 Kelvin.Here the symbol ^ represents power.Please help me in solving this problem.

Thank you sir. I have solved it with your help.
 
why can a sphere be regarded as a point source?
 
kingkong said:
why can a sphere be regarded as a point source?
What is the context for this ? In most cases, the answer is in Gauss' Law.
 
i am doing an experiment invesitigating the inverse square law of radiation using a bulb. a bulb is considered as a point source thus, a sphere. Please help.
 
kingkong said:
i am doing an experiment invesitigating the inverse square law of radiation using a bulb. a bulb is considered as a point source thus, a sphere. Please help.
If this is an uncoated, incandescent bulb, then you are treating it as a point source because all the light comes from the tiny length of filament inside it. If this length is small comapred to the distances at which you are measuring the intensity, then it can be treated as a "point". This is an approximation, but usually a good one. And this has little to do with the fact the the bulb is shaped like a sphere - in fact, most bulbs aren't.

However, if you did have a spherical body (like a planet or star or spherical phosphorescent bulb), then too, you can treat the radiation as coming from a point source centered with the body. The reason for this is that the radiation from the body mostly comes out in directions normal to the surface.

If instead of a spherical body, you had a long, cylindrical body, then by the same argument, you could treat it as a line-source.
 
thank you very much
 

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