An electron of mass m, when accelerated through a potential V has de-Broglie wavelength λ the de-Broglie wavelength associated 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 electron, pe=(2meV)Momentum of electron pp=(2MeV)λpλe=h/peh/pe=pepp=2meV2MeV=mM λp=λemM=λmM