A proton and an alpha – particle are accelerated through same potential difference.
As we know,
1/2 mv2 = qV
mv = √(2mqV)
Since, wavelenegth is given by:
Now, the ratio of de-Broglie wavelength of proton and alpha-particle will be given by:
λp/λα = (√qm)α/(√qm)p
= √(2 x 4)/√(1 x 1)