The French physicist Louis de Brogie in 1924 postulated that matter, like radiation, should exhibit a dual behaviour. He proposed the following relationship between the wavelength `lambda` of a material particle, its linear momentum p and planck constant h.
`lambda =(h)/(p) = (h)/(mv)`
The de Broglie relation implies that the wavelength of a particle should decrease as its velocity increase. its also implies that for a given velocity heavier particles should have shorter wavelength than lighter particles. The waves associated with particles in motion are called matter waves or de Broglie waves. These waves differe from the electromagnetic waves as they
(i) have lower velocites
(ii) have no electrcal and magnetic fields, and
(iii) are not emitted by the particle under consideration. The experimental confirmation of the de Brgolie's relation was obtained when Davisson and Germer, in 1927, observed that a beam of electron is diffracted by a nickel crystal. As diffraction is a characterstic property of waves hence the beam of electron behaves as a wave, as proposed by de Broglie
If proton, electron and `alpha`-particle are moving with same kinetic energy then the order of their de-Broglie's wavelength.

Options

A. `lambda_P gt lambda_e gt lambda_alpha`
B. `lambda_alpha gt lambda_P gt lambda_e`
C. `lambda_alpha lt lambda_P lt lambda_e`
D. `lambda_e = lambda_P lt lambda_alpha`

Answer

C

Explanation

`lambda=h/sqrt(2mKE)` if KE same `lambda prop 1/sqrtm` `m uarr` `lambdadarr`