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Questions
The de Broglie wavelengths associated with an electron and a proton are the same. What will be the ratio of
- their momenta
- their kinetic energies?
The de-Broglie wavelengths associated with an electron and a proton are same. Calculate the ratio of their kinetic energies. (Given : mp = 1836 me)
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Solution
Data: λ (electron) = λ (proton)
m (proton) = 1836 m (electron)
(i) λ = `h/p` As λ (electron) = λ (proton),
`(p("electron"))/(p("proton")) = 1`, where p denotes the magnitude of momentum.
(ii) Assuming v < < c,
KE = `1/2mv^2 = 1/2 (m^2 v^2)/m = p^2/(2m)`
`(KE("electron"))/(KE("proton")) = ((p_"electron")/(p_"proton"))^2 . (m_("proton"))/(m_("electron"))`
= 1 × 1836
= 1836 as p is the same for the electron and the proton.
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