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Question
A proton and an α-particle are accelerated, using the same potential difference. How are the de-Broglie wavelengths λp and λa related to each other?
Solution
In this problem since both proton and α-particle are accelerated through the same potential difference,
We know that, `λ = h/sqrt(2mqv)`
∴ `λ oo 1/sqrt(mq)`
`λ_p/λ_α = sqrt(m_αq_α)/(m_pq_P) = sqrt(4m_p xx 2e)/sqrt(m_p xx e) = sqrt(8)`
∴ `λ_p = sqrt(8)λ_α`
i.e., wavelength of proton is `sqrt(8)` times wavelength of α-particle.
Important point: De-Broglie wavelength associated with the charged particles: The energy of a charged particle accelerated through potential difference V is `E = 1/2 mv^2 = qV`
Hence de-Broglie wavelength `λ = h/p = h/sqrt(2mE) = h/sqrt(2mqV)`
`λ_("Electron") = 12.27/sqrt(V) Å, λ_("Proton") = 0.286/sqrt(V) Å`
`λ_("Deutron") = 0.202/sqrt(V) Å, λ_(a"-particle") = 0.101/sqrt(V) Å`
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