Compton and de Broglie

Classical Mechanics Level 2

For a non-relativistic particle, which of the following gives the correct relationship between the de Broglie wavelength $\lambda_{dB}$ and the Compton wavelength $\lambda_c = \frac{h}{mc}$ , which is also a useful wavelength in quantum mechanics?

\lambda_{dB} > \lambda_c

Not enough information.

\lambda_{dB} < \lambda_c

\lambda_{dB} = \lambda_c

1 solution

Matt DeCross
Apr 23, 2016

The de Broglie wavelength is:

$\lambda_{dB} = \frac{h}{p}.$

For a non-relativistic particle, $p = mv$ and $v \ll c$ , so:

$\lambda_{dB} \approx \frac{h}{mv} > \frac{h}{mc} = \lambda_c$

Thus $\lambda_{dB} > \lambda_c$ as claimed.

Compton and de Broglie

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