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A proton, a neutron, an electron and an `alpha`-particle have same energy. Then their de-Broglie wavelengths compare asA. `lambda_(p)=lambda_(n)gtlambda_(e)gtlambda_(alpha)`B. `lambda_(alpha) lt lambda_(p)=lambda_(n)gtlambda_(e)`C. `lambda_(e)ltlambda_(p)=lambda_(n)gtlambda_(alpha)`D. `lambda_(e)=lambda_(p)=lambda_(n)=lambda_(alpha)` |
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Answer» Correct Answer - B We know that the relation between `lambda` and `K` is given by `lambda=h/sqrt(2mk)` Here, for the given value of energy `K,h/sqrt(2k)` is a constant. Thus, `lambda prop 1sqrtm` `therefore lambda_(p):lambda_(n):lambda_(e):lambda_(alpha)` `1/sqrtm_(p):1/sqrtm_(n):1/sqrtm_(e):1/sqrtm_(alpha)` Since, `m_(p)=m_(n)`, hence `lambda_(p)=lambda_(alpha)` As,`m_(alpha) gt m_(p)`, therefore `lambda_(alpha) lt lambda_(p)` As,`m_(e) lt m_(n)`, therefore `lambda_(e) gt lambda_(n)` Hence,`lambda_(alpha) lt lambda_(p)=lambda_(n) lt lambda_(e)` |
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