Question 10: Dual Nature of Matter and Radiation

Question

de Broglie wave length associated with an electron accelerated through a potential difference ' V ' is ' $\lambda$ '. If the accelerating potential is halved, what will be the new wave length associated with the charged particle?

Accepted Answer

The de Broglie wavelength for an electron accelerated through a potential $V$ is
$$ \lambda \;=\;\frac{h}{p} \;=\;\frac{h}{\sqrt{2meV}} \;\propto\;V^{-1/2}. $$
If we halve the potential $\displaystyle V	o 	frac V2$, then
$$ \lambda_{m new} =\frac{h}{\sqrt{2me\,(V/2)}} =\frac{h}{\sqrt{	frac12\,2meV}} =\sqrt{2}\,\frac{h}{\sqrt{2meV}} =\sqrt{2}\,\lambda. $$
So the new wavelength is $\displaystyle \sqrt{2}\,\lambda$, i.e. Option D.

Answer

$\frac{\lambda}{\sqrt{2}}$

Answer

$\frac{\lambda}{2}$

Answer

$2 \lambda$

Answer

$\sqrt{2} \boldsymbol{\lambda}$

Dual Nature of Matter and Radiation

de Broglie wave length associated with an electron accelerated through a potential difference ' V ' is ' \lambda '. If the accelerating potential is halved, what will be the new wave length associated with the charged particle?

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