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An $α$ particle and proton are accelerated from rest by a potential difference of 200 V. After this, their de Broglie wavelengths are $λ_{α}$ and $λ_{p}$ respectively. The ratio $\frac{λ_{p}}{λ_{α}}$ is:

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Correct option is A

Kinetic energy K=P22m=qV⇒P=2mqV⇒P ∝ mqNow, λ=hP⇒λ∝1P⇒λpλα=PαPp=mαqαmpqp=4×21×1 ⇒λpλα=22=2.8

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An α particle and proton are accelerated from rest by a potential difference of 200 V. After this, their de Broglie wavelengths are λα and λp respectively. The ratio λpλα is:

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An $α$ particle and proton are accelerated from rest by a potential difference of 200 V. After this, their de Broglie wavelengths are $λ_{α}$ and $λ_{p}$ respectively. The ratio $\frac{λ_{p}}{λ_{α}}$ is: