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

Explanation

For a particle, de-Broglie wave length,`lambda = h//p`
Kinetic energy,`K=p^(2)//2m`
Then, `lambda=(h)/(sqrt(2mK))`
For the same kinetic energy K, the de-Broglie wave length associated with the particle is inversely proportional to the square root of their masses. A proton (`""_(1)^(1)H)` is 1836 times massive than an electron and `alpha`-particle (`""_(4)^(2)H)` four times that of a proton. Hence, `alpha`-particle has the shortest de-Broglie wave length.
`lambda=(h)/(p)=(h)/(mv)`
Mass `m=(h)/(lambda v)`
For an electron, mass `m_(e)=(h)/(lambda_(e)v_(e))`
`=1.813xx10^(-4)`
Then, mass of the particle `m=m_(e)((lambda_(e))/(lambda))((v_(e))/(v))`
`m=(9.11 xx 10^(31 kg) xx (1)/(3) xx ((1)/(1.813 xx 10^(-4)))`
`m=1.675 xx 10^(-27) kg`.
Thus, the particle, with this mass could be a proton or a neutron.