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