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