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A proton accelerated by potential difference 100 V have de-Broglie wavelength of lambda_(0) alpha-particle is accelerated by same potential different.Its de-Broglie wavelength will be …… |
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Answer» `2sqrt(2)lambda_(0)` `THEREFORE m^(2)v^(2)=p^(2)=2meV` Now `LAMBDA=(h)/(p)` `therefore lambda=(h)/(sqrt(2meV))` `therefore m^(2)v^(2)=p^(2)=2meV` Now `lambda=(h)/(p)` `therefore lambda=(h)/(sqrt(2meV))` `therefore lambda prop (1)/(sqrt(me))` `therefore (lambda_(alpha))/(lambda_(p))=sqrt((m_(p))/(m_(alpha)e_(alpha)))` Now `m_(alpha)=4m_(p),e_(alpha)=2e_(p)`, `therefore (lambda_(a))/(lambda_(p))=sqrt((m_(p)e_(p))/(4m_(p)xx2e_(p)))=(1)/(sqrt(8))` `therefore (lambda_(alpha))/(lambda_(p))=(1)/(2sqrt(2))` `therefore (lambda_(alpha))/(lambda_(0))=(1)/(2sqrt(2)) therefore lambda_(alpha)=(lambda_(0))/(2sqrt(2))` |
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