de broglie wavelength

Question

An electron is accelerated through a potential difference of 100 V. Its de Broglie wavelength is approximately:

RankGuru · Accepted Answer

The **de Broglie Wavelength ($\lambda$)** of an accelerated electron is given by:
$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2meV}}$
Using standard values for $h, m, e$, the approximate formula is:
$\lambda \approx \sqrt{\frac{150}{V}} 	ext{ \AA} \approx \frac{12.27}{\sqrt{V}} 	ext{ \AA}$
Given $V = 100 	ext{ V}$:
$\lambda = \frac{12.27}{\sqrt{100}} = \frac{12.27}{10} = 1.227 	ext{ \AA}$
Since $1 	ext{ nm} = 10 	ext{ \AA}$:
$\lambda \approx 0.1227 	ext{ nm} \approx 0.123 	ext{ nm}$

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de broglie wavelength

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