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.

Explanation

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.