get de-broglie wavelength of thermal neutron at 27celcius

Asked by Ashok Rana | 8th Mar, 2015, 09:00: AM

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 Faiza Lambe | 8th Mar, 2015, 09:47: AM