Question
Sun March 08, 2015 By: Ashok Rana

get de-broglie wavelength of thermal neutron at 27celcius

Expert Reply
Faiza Lambe
Sun March 08, 2015
begin mathsize 14px style Energy space of space thermal space neutrons space at space ordinary space temperatures space straight E space equals space kT
therefore straight lambda equals fraction numerator straight h over denominator square root of 2 mkT end root end fraction equals fraction numerator 30.835 straight A to the power of straight o over denominator square root of straight T end fraction
The space De space Broglie space wavelength space of space thermal space neutrons comma
straight lambda subscript straight n equals fraction numerator 30.8 straight A to the power of straight o over denominator square root of straight T end fraction equals fraction numerator 30.8 over denominator square root of 300 end fraction equals 1.77 straight A to the power of straight o end style
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