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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