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An α -particle and a proton are accelerated from rest by a potential difference of 400V. After this, their de- Broglie wavelengths are λα and λp respectively. The ratioλpλα is equal to

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a

2

b

2√2

c

4

d

4√2

answer is B.

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Detailed Solution

λ=h2mqV⇒λ∝1mq⇒λpλα=mαqαmpqp=4mm2ee=8=22

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