Question

An electron and an alpha particle have the same kinetic energy. How are the de Broglie wavelengths associated with them related?

Answer 1

λ = `"h"/"p"`

Kinetic energy of the particle K = `1/2 "mv"^2 = "P"^2/(2"m") = "h"/(2"m"λ^2)`

i.e. λ = `"h"/sqrt(2"mK")`; `λ ∝ 1/sqrt"m"`

`λ_"e"/λ_"α" = sqrt("m"_α/"m"_"e")`