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

Explanation

K.E. ,
`K=1/2 (mv^2)/m=1/2 (p^2)/m`
or `p=sqrt2mk` ,
De-Broglie wavelength,
`lambda=h/p=h/(sqrt(2mK)) or lambdaprop 1/(sqrtm)`
Since proton is 1836 times heavier than electron and `alpha`-particle is 4 times heavier than proton so,
`m_e:m_p:m_(alpha)=1:1836:(1836xx4)`
`:. lambda` is shortest for `alpha`-particle.