Find the ratio of de Broglie wavelength of a proton and an alpha-particle which have been accelerated through same potential difference.

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22:1

3:2

32:1

2:1

answer is A.

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Detailed Solution

de Broglie wavelength =λ=hmv=h2mqV∴λpλα=mαqαVαmpqpVp Putting Vα=Vp,⇒λpλα=4×21×1=22

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