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A proton and an electron are accelerated by the same potential difference. Let $λ_{e}$ and $λ_{p}$ denote the de Broglie wavelength of the electron and the proton respectively, then
$λ_{e} = λ_{p}$
$λ_{e} < λ_{p}$
$λ_{e} > λ_{p}$
$λ_{e} \geq λ_{p}$

A

λ_{e} \geq λ_{p}

B

λ_{e} = λ_{p}

C

λ_{e} < λ_{p}

D

λ_{e} > λ_{p}

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Solution

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$\frac{1}{2} m v^{2} = e V$

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