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Two particles A and B of de-Broglie wavelength lambda_(1) and lambda_(2) combine to form a particle C.The process conserves momentum .Find the de-Broglie wavelength of the particle C.(The motion is one -dimensional). |
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Answer» <P> SOLUTION :Here `vecPA||vecPB`,hence `P_(C)=p_(A)+p_(B)`….(1)There are four possiblities: (i)`p_(B)gt0,p_(A)ggt0` both are positive . `THEREFORE (h)/(lambda_(c))=(h)/(lambda_(A))+(h)/(lambda_(B))IMPLIES(1)/(lambdaC)=(1)/(lambdaA)+(1)/(lambda_(B))` `therefore lambda_(C )=(lambda_(A)lambda_(B))/(lambda_(A)lambda_(B))` (ii)`p_(A)lt0,p_(B)lt0` both are negative. `therefore (h)/(lambdac)=(-h)/((lambda_(A)))+(-h)/(lambda_(B))` `therefore lamba_(C)=-((lambda_(A)lambda_(B))/(lambda_(A)+lambda_(B)))` (iii)`p_(A)gt0,p_(B)lt0p_(A)` ispositive and `p_(B)` is negative `therefore (h)/(lambda_(C))=(h)/(lambda_(A))-(h)/(lambda_(B))implies(1)/(lambda_(C))=(1)/(lambda_(A))-(1)/(lambda_(B))` `therefore lambda_(C)=(lambda_(A)lambda_(B))/(lambda_(B)-lambda_(A))` (iv)`p_(A)lt0,p_(B)gt0p_(A)` is positive and `p_(B)` is negative `(h)/(lambda_(C))=(-h)/(lambda_(A))+(h)/(lambda_(B))implies(1)/(lambda_(C))=(1)/(lambda_(B))=(1)/(lambda_(A))` `therefore lambda_(C)=(lambda_(A)lambda_(B))/(lambda_(A)-lambda_(B))` |
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