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An electron ,an alpha-particle and a proton have the same kinetic energy .Which of these particles has the shortest de-Broglie wavelength? |
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Answer» Solution :Mass of electron `m_(E)`,mass of proton `m_(p)`, and mass of `alpha`-particle `m_(alpha)` `therefore m_(e)lt m_(p)ltm_(alpha)` `implies` Here KINETIC energy of all three is equal and relation between kinetic energy and momentum, `K=(p^(2))/(2m) therefore p=sqrt(2mK)` `implies` Now,the de-Broglie WAVELENGTH of particle, In `LAMBDA=(h)/(p)=(h)/(sqrt(2mK))`, h,2,k are equal `therefore lambda prop (1)/(sqrt(m))` Here mass of `alpha`-particle is maximum hence de-Broglie wavelength of `alpha` particle will be minimum. |
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