A proton and an -particle are accelerated from rest to the same energy. The de-Broglie wavelengths _p and _ are in the ratio

Step-by-step solution

The de-Broglie wavelength associated with a charged particle is given by = h p where, h = planck's constant and p = momentum = 2 mKE (here, KE is the kinetic energy of the charged particle) = h 2 mKE For proton and -particle, the wavelengths are respectively given as, aligned _ p = h 2 ~m _ p KE _ p and _ = h 2 ~m _ KE _ _ p _ = 2 ~m _ KE _ 2 ~m _ p KE _ P aligned Here, KE _ = KE _ p and m _ =4 ~m _ p Substituting these above mentioned relations in Eq. (i), we get _ P _ = 4 ~m _ P m _ p = 2 1 or _ P : _ =2: 1