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Question
An electron, an alpha particle and a proton have the same kinetic energy.
Which one of these particles has the largest de-Broglie wavelength?
Solution
De-Broglie’s wavelength is given as, λ = `h/(sqrt(2mK))`
Where, h is Planck’s constant, m is the mass of the particle, K is the kinetic energy of the particle
The mass of the electron is least among an electron, an alpha particle, and a photon.
`because lambda ∝ 1/sqrtm`, therefore, the electron has the largest de-Broglie wavelength.
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