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An `alpha`-particle when accelerated through a potential of V volt has a wavelength `lambda` associated with it, but if a proton in order to have same wavelength `lambda` by what potential difference it must be accelerated?A. 8 VB. 6 VC. 4 VD. 12 V |
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Answer» Correct Answer - A `lambda_(p)=lambda_(alpha)` `(h)/(sqrt(2m_(p)Q_(p)V_(p)))=(h)/(sqrt(2m_(alpha)Q_(alpha)V_(alpha)))` `therefore" "m_(p)Q_(p)V_(p)=m_(alpha)Q_(alpha)V_(alpha)` `therefore" "V_(p)=((m_(alpha))/(m_(p)))((Q_(alpha))/(Q_(p)))V_(alpha` `V_(p)=(4)(2)V_(alpha)implies V_(p)=8V` |
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