An electron of mass m whet accelerated through a potential difference V has de-Broglie wavelength λ The de-Broglie wavelength associated with a proton of mass M accelerated through the same potential difference will be
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a
λmM
b
λmM
c
λMm
d
λMm
answer is B.
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Detailed Solution
Momentum of electrons Pe=2meVMomentum for proton Pp=2MeVλpλe=hPphPe=PePp=2meV2MeV λpλe=mM λp=λemM∴(λe=λ) λp=λmM