Practice portal › Dual Nature of Matter and Radiation › Comparing de Broglie Wavelengths of Particles
Asked in GUJCET 2022 · Same or changed energy
Given: the same kinetic energy K for all three particles.
Idea: p = √2mK, so λ = h/(√2mK) ∝ 1/(√m) at fixed K.
The heaviest particle has the largest momentum and so the shortest wavelength.
Masses: mₑ < mₚ < m_α, with m_α ≈ 4mₚ.
So the α-particle has the shortest de Broglie wavelength.
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