Dual Nature And Uncertainty

Question

A proton and an alpha particle are both accelerated through the same potential difference V. What is the ratio of their de Broglie wavelengths λ_proton / λ_alpha, given that m_alpha = 4m_proton and charge on alpha particle is 2e?

RankGuru · Accepted Answer

When a charged particle of charge q and mass m is accelerated through potential difference V, it gains kinetic energy qV = p^2/2m, giving momentum p = √(2mqV). The de Broglie wavelength is λ = h/p = h/√(2mqV). For the proton: λ_p = h/√(2m_p e V). For the alpha particle: λ_α = h/√(2 × 4m_p × 2e × V) = h/√(16 m_p e V). The ratio: λ_p/λ_α = [h/√(2m_p eV)] / [h/√(16m_p eV)] = √(16m_p eV) / √(2m_p eV) = √(16/2) = √8 = 2√2. Therefore λ_proton : λ_alpha = 2√2 : 1. Option 1:√2 corresponds to an error in ignoring the charge difference (using q=e for both). Option √2:1 corresponds to considering only the mass ratio and not the charge ratio. Option 1:2√2 is the inverse of the correct answer, confusing the numerator and denominator. This problem tests de Broglie wavelength in the context of charged particle acceleration, a classic JEE Advanced topic combining electrostatics with wave-particle duality. Plausibility check: the alpha particle has both larger mass and larger charge than the proton; both factors increase its momentum, reducing its wavelength. So λ_proton > λ_alpha, meaning the ratio λ_p/λ_α > 1, and 2√2 > 1 is consistent with this physical expectation.

Dual Nature And Uncertainty

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