An electron and a proton, each have de Broglie wavelength of 1.00 nm.
Compare the kinetic energy of the proton with that of the electron.

Explanation

From relation `K = (1)/(2) m upsilon^(2) = ( p^(2))/(2m)`
`K_(e ) = ( p_( e)^(2))/( 2m_(e )) ` and `K_(p) = ( p_(p)^(2))/(2m_(p))`
`(K_(p))/( K_( e)) = ( p_(p)^(2))/( 2m_(p)) xx ( 2m_(e))/( p_(e)^(2)) = ( m_(e ))/( m_(p))`
Since` m_(e ) lt lt lt m_(P) `, So `K_(p ) lt lt lt K_(e ) `
`(K_(p))/( K_(e )) = ( 9.1 xx 10^(-31))/( 1.67 xx 10^(-27))`
`= 5.4 xx 10^(-4)`