An electron and proton are possessing the same amount of kinetic energy. Which of the two has greater wavelength?

Explanation

`K=(1)/(2)mv^(2)=(p^(2))/(2m), lambda=(h)/(p), p=(h)/(lambda), therefore K=(h^(2))/(2m lambda^(2))`
KE of electron, `K_(e )=(h^(2))/(2m_(e )lambda_(e )^(2))`
KE of proton, `K_(p)=(h^(2))/(2m_(p)lambda_(p)^(2))`
Given, `K_( e)=K_(p)`
`(h^(2))/(2m_(e )lambda_(e )^(2))=(h^(2))/(2m_(p)lambda_(p)^(2))`
`(lambda_(e ))/(lambda_(p))=sqrt((m_(p))/(m_(e )))`
`m_(p) gt m_(e )`, Hence, `lambda_(e ) gt lambda_(p)`.