Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12

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

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)

Concept: de-Broglie Relation
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