First slide

Matter waves (DC Broglie waves)

Question

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

Easy

A

$λ \sqrt{(\frac{M}{m})}$
B

$λ \sqrt{(\frac{m}{M})}$
C

$λ (\frac{M}{m})$
D

$λ (\frac{m}{M})$

Solution

Momentum of electron, $p_{e} = \sqrt{(2 meV)}$
Momentum of electron $p_{p} = \sqrt{(2 MeV)}$
$\frac{λ_{p}}{λ_{e}} = \frac{h / p_{e}}{h / p_{e}} = \frac{p_{e}}{p_{p}} = \frac{\sqrt{2 meV}}{\sqrt{2 MeV}} = \sqrt{(\frac{m}{M})} λ_{p} = λ_{e} \sqrt{(\frac{m}{M})} = λ \sqrt{(\frac{m}{M})}$

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