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Q.

When a proton is accelerated through a potential difference of 4000 volt, its de Broglie wavelength is λ. What will be the de Broglie wavelength of a deuteron when accelerated through a potential difference of 8000 volt?

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a

λ2

b

2λ

c

λ22

d

λ2

answer is D.

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Detailed Solution

p22m=Q.V⇒p=2mQV∴ λ=hp=h2mQV∴ λdλp=mpmdQpQd.VpVd=12.1.40008000=12⇒λd=λ2
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