A proton accelerated through a potential difference of 100V, has de – Brogile wavelength λ0 . The de – Brogile wavelength of an α - particle, accelerated through 800 V is

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λ02

λ04

λ08

answer is D.

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Detailed Solution

p22m=qVp=2mqVλ=h2mqVλ1λ2=m2q2V2m1q1V1=4m2q800mq100λ0λ2=64λ2=λo8

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