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A proton and an electron have same de Broglie wavelength. Which of them moves faster and which possesses more kinetic energy ? Justify your answer. |
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Answer» Solution :We KNOW that `lambda=(H)/sqrt(2mK)` Since proton and electron have same de Broglie wavelength, we get `(h)/sqrt(2m_(p) K_(p))=(h)/sqrt(2m_(e)K_(e))` Since `m_(e) lt m_(p). K_(p) lt K_(e)`, the electron has more kinetic ENERGY than the proton. `K_(p)/K_(e)=(1/2 m_(p) v_(p)^(2))/(1/2 m_(e)v_(e)^(2))` `v_(p)/v_(e)=sqrt((m_(e)^(2))/(m_(p)^(2)) =m_(e)/m_(p)" since "K_(p)/K_(e)=m_(e)/m_(p)`, Since `m_(e) lt m_(p), v_(p) lt v_(e)`, the electron moves FASTER than the proton. |
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