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A proton and an alpha-particle have the same de-Broglie wavelength. Determine the ratio of (i) their accelerating potentials (ii) their speeds. |
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Answer» <P> Solution :As PER QUESTION `lamda_(p)=lamda_(alpha)` but `lamda=(h)/(sqrt(2qmV))=(h)/(mv)`(i) `THEREFORE (h)/(sqrt(2em_(p)V_(p)))=(h)/(sqrt(2(2e)(4m_(p))V_(alpha)))implies 2em_(p)V_(p)=16em_(p)V_(alpha)implies(V_(p))/(V_(alpha))=(8)/(1)` (ii) Again `lamda=(h)/(m_(p)v_(p))=(h)/(4m_(p)v_(alpha))implies m_(p)v_(p)=4m_(p)v_(alpha)implies(v_(p))/(v_(alpha))=4/1`. |
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