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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)
The de-Broglie wavelengths associated with an electron and proton are same. Calculate the ratio of their momentum and kinetic energies.
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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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