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A particle is moving three times as Fast as an electron .The ratio of the de-Broglie wavelength of the particle to that of the electron is 1.813xx10^(-4). Calculate the particle's mass and indentify the particle. |
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Answer» Solution :Suppose ,the speed of ELECTRON is `v_(e)` and speed of PARTICLE is v.Mass of electron `m_(e)` and mass of particle is m and de-Broglie wavelength of electron `lambda_(e)` and de-Broglie wavelength of particle is `lambda`. Now de-Broglie wavelength of particle, `lambda=(H)/(p)=(h)/(mv)` [`thereforep=mv`] `therefore m=(h)/(lambdav)` `therefore` For electron `m_(e)=(h)/(lambda_(e)v_(e))` For particle `m=(h)/(lambdav)` `therefore (m)/(m_(e))=(lambda_(e))/(lambda)XX(v_(e))/(v)` `therefore m=(m_(e))xx(lambda_(e))/(lambda)xx(v_(e))/(v)` `=9.11xx10^(-31)xx(1)/(1.813xx10^(-4))xx(1)/(3)` `therefore m=1.675xx10^(-27)kg` `therefore` This particle can be proton or NEUTRON. |
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