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A narrow stream of monoenrgetic electrons falls at an angle of incidence theta=30^(@) on the natural facet of a n aluminum single crystal. The distacne between the neighbouring cystal planesparallel to that fact to d= 0.20nm. Theaximum mirror reflection is observed at a certain accelecrating voltage V_(0). Find V_(0) if the next maximum mirror reflection is known to be observed when the acceleration voltage is increased eta=2.25 time s |
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Answer» Solution :From Bragg's law, for the first case `2 d sin THETA=n_(0)lambda=n_(0)(2pi ħ)/(sqrt(m e etaV_(0)))` THUS `n_(0)=(n_(0)+1)/(sqrt(eta))` or `n_(0)(1-(1)/(sqrt(eta)))=(1)/(sqrt(eta))` or `n_(0)=(1)/(sqrt(eta-1))` Going back we get `V_(0)=(pi^(2) ħ^(2))/(2 me d^(2)sin^(2) theta).(1)/((sqrt(eta-1))^(2))- 0.150keV` Note: In the Bragg's formula, `theta` is the glancing angle and not the incidence. We have obtained correct RESULT by taking `theta` to be the glancing angle. If `theta` is the angle of incidence, then the glancingk angle will be `90-theta`. Then the final ANSWER will be similar by a factor `TAN^(2)theta=(1)/(3)`. |
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