An electron ,an `alpha`-particle and a proton have the same kinetic energy .Which of these particles has the shortest de-Broglie wavelength?

Explanation

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.