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Home / Doubts and Solutions/JEE/Physics

 AN ELECTRON BEAM OF ENERGY 10 KEV IS INCIDENT ON METALLIC FOIL IF THE INTERATOMIC DISTANCE IS 0.55 ANGSTROM FIND THE ANGLE OG DIFFRACTION.
 
 
 
 
PLS GIVE THE ANS FAST

Asked by sudheerkapoor67 9th October 2017, 6:27 PM
Answered by Expert
Answer:
Energy of electron E = 10 keV = 10 × 1000 × 1.6 ×10-19 J 
 
de Broglie's wave length = λ = begin mathsize 12px style fraction numerator straight h over denominator square root of 2 mE end root end fraction end style = begin mathsize 12px style fraction numerator 6.63 space cross times space 10 to the power of negative 34 end exponent over denominator square root of 2 space cross times space 9.1 space cross times 10 to the power of negative 31 end exponent cross times 10 cross times 1000 cross times 1.6 cross times 10 to the power of negative 19 end exponent end root end fraction end style = 1.229 x 10-11 m
 
Bragg's law 2d sin θ = nλ
 
first order diffraction n = 1
 
hence sin θ = begin mathsize 12px style fraction numerator n lambda over denominator 2 d end fraction end style = begin mathsize 12px style fraction numerator 1.229 space cross times 10 to the power of negative 11 end exponent over denominator 2 space cross times space 0.55 space cross times space 10 to the power of negative 10 end exponent end fraction space equals space 0.1117 end style
 
hence θ = 6.415 °
Answered by Expert 11th October 2017, 11:47 AM
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