Chemical Forums
Chemistry Forums for Students => Inorganic Chemistry Forum => Topic started by: tashkent on November 20, 2004, 01:14:26 PM
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Greetings everyone!
I'd like to know the principle on how the xenon lamp works? I encountered this confusing question in the GRE Chemistry Exam recently. It has something to do with the interaction of Xenon and Fluorine, forming XeF6. Some of the choices were: Xenon photodissociates from ground to excited state; Fluorine photodissociates from ground to excited state...I forgot the other three choices. I will appreciate any explanation from u guys. My friend told me that fluorine should photodissociate, since its p orbitals are not completely filled...
Hope u can shed some light on this matter. Thank you.
Regards,
Tashkent
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Wah! Inorganic chemists are very quiet about this topic :-\
Anyway, I'm sad coz I've read from Atkins that XeF2 is formed in a glass bulb where xenon and fluorine are present and then exposed to sunlight. Fluorine undergoes photodissociation , then photochemically generated F atoms react with Xe atoms. I should've thought about this, since xenon cannot photodissociate anymore, due to its noble gas configuration...I'm wondering if that's right...
Oh well, I'll just shed a tear coz I haven't thought of this during the GRE Chemistry exam :'(
Listening to Expose's "I'll Never Get Over You",
Tashkent
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You'll learn about xenon lamps in analytical/instrumental chem. They are usually used in spectrofluoroeters where a source of continuum radiation is needed. The spectrum is I think about a few hundred to 1300nm and emists blackbody radiation, I think...