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Question
A proton and an electron have same kinetic. Which one has greater de-Broglie wavelength and why?
Solution
Since de-Broglie wavelength λ in terms of kinetic energy is given as
`lambda = h/(sqrt(2mE_k))`
Where Ek is kinetic energy, m is mass of electron and h is the Planck’s constant.
Thus for electron and proton with same kinetic energy, de Broglie wavelength would depend on mass.
Since `lambda ∝= 1/sqrtm`
Now mp > me so, λe>λp.
Hence wavelength of electron is greater than wavelength of proto
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