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Question

A proton and electron are accelerated by same potential difference starting from the rest have de-Broglie wavelength λp and λe.

A
λe=λp
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B
λe<λp
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C
λe>λp
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D
none of these
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Solution

The correct option is C λe>λp
Let me and mp be the mass of electron and proton respectively. Let the applied potential different be V.
Thus, the de-Broglie wavelength of the e,
λe=h2meeV(1)
and de - Broglie wavelength of proton
λp=h2mpeV(2)
Dividing,
λpλe=memp
me<mp
λpλe<1
λpλe

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