An α-particle and a proton are accelerated through the same potential difference. Find the ratio of their de Broglie wavelength.

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#### Solution

We know that,

Charge on the proton = e

Charge on the alpha particle = 2e

Let mass of proton = m

So, mass of alpha particle = 4m

When a particle is of mass m and change q is accelerated by a potential V, then its de Broglie wavelength is given by

`λ=h/sqrt(2mqV)`

For proton:

`λ_"proton"=h/sqrt(2meV)`

For alpha particle:

`λ_"alpha particle"=h/sqrt(2×4m×2e×V)`

`⇒λ_"alpha particle"/λ_"proton"=(h/(2×4m×2e×V))/(h/(2meV))`

`⇒λ_"alpha particle"/λ_"proton"=(2meV)/(2×4m×2e×V)`

`=1/(2sqrt2)`

Concept: de-Broglie Relation

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