Answer: (C)2√2

Explanation:

Given,

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:

λ= h/mv

λ=h/√(2mqV)

λ∝1/√(qm)

Now, the ratio of de-Broglie wavelength of proton and alpha-particle will be given by:

λpα = (√qm)α/(√qm)p

= √(2 x 4)/√(1 x 1)

= 2√2

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