λρ > λα > λe
λe > λρ > λα
λα > λe > λρ
λα > λρ > λe
For a given kinetic energy K, the de Broglie wavelength linked to a particle displays an inverse relationship with the square root of its mass. Comparing a proton (1H), which is 1836 times more massive than an electron, and an alpha particle (4He), which is four times heavier than a proton, it becomes evident that the alpha particle possesses the shortest de Broglie wavelength due to its relatively larger mass.
∴ μe < mρ < mα
∴ λe > λρ > λα
The correct option is (B): λe > λρ > λα
If the momentum of an electron is changed by P, then the de Broglie wavelength associated with it changes by \(1\%\). The initial momentum of the electron will be:
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