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If proton and electron have same de-Broglie wavelength then…. |
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Answer» both will have same kinetic energy Here `lambda` is same (GIVEN) `THEREFORE (h)/(m_(p)v_(p))=(h)/(m_(e)v_(e))` `therefore m_(p)v_(p)=m_(e)v_(e)` `(v_(p))/(v_(e))=(m_(e))/(m_(p))` `therefore (E_(p))/(E_(e))=((1)/(2)m_(p)vp^(2))/((1)/(2)m_(2)v_(e)^(2))` `=(m_(p))/(m_(e))xx((m_(e))/(m_(p)))^(2)=(m_(e))/(m_(p))` but `m_(e)ltm_(p)` thus `(m_(e))/(m_(p))lt1` `therefore (E_(P))/(E_(e))lt1 therefore E_(P)ltE_(e)` |
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