An alpha particle and a proton are accelerated from rest through the same potential difference V. Find the ratio of de-Broglie wavelengths associated with them.

The de-Broglie wavelength associated witha proton having charge e and accelerated through a potential difference V is given by

λ = h /( 2mev)1/2 ......... (i)

suppose that α - particle posses the de-Broglie wavelength λ' , when accelerated through the same V. if e' amd m' are the values of charge and mass of the α -particle , then

λ' = h/(2m'e' V)1/2 ......... (ii)

dividing eqn (i) by (ii) , we have

λ / λ' = h / (2meV)1/2 x (2m'e' V)1/2 / h

= (m' e' / m e)1/2

as u know , mass of alpha particle is 4 times the proton and charge is 2 times greater,

m' = 4m ; e' = 2e

therefore

λ / λ ' = (4m x 2e / m e)1/2

= (8)1/2

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