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get de-broglie wavelength of thermal neutron at 27celcius

Asked by Ashok Rana 8th March 2015, 9:00 AM
Answered by Expert
Answer:
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
Answered by Expert 8th March 2015, 9:47 AM
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