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For a chemistry problem, I want the velocity of the de Broglie wave to agree with one of these values:

A) $c^2/v$
B) $h\nu/mc$
C) $mc^2/h$
D) $v\lambda$

We know that $$\lambda = \frac{h} {mv}$$

But I am confused by the next step in my book:

$$\lambda = \frac {c} {\nu}$$

After which:

\begin{align} \frac{h}{mv} &= \frac{c}{\nu} \\ \implies v &= \frac{h\nu}{mc} \end{align}

Why is the de Broglie wavelength $\lambda = c/\nu$? In every context I've seen it used, $c$ is the speed of light. But I don't think the velocity of the matter wave could equal the speed of light, because the matter wave is only as fast as the particle – which, having mass, cannot travel at the speed of light.

What am I doing wrong here? Does $c$ moonlight in quantum chemistry as the velocity of the matter wave, separate from its connotations in physics?

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  • $\begingroup$ Read this: Matter Waves $\endgroup$ – Firefox1921 Jun 18 at 9:56
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    $\begingroup$ Welcome to Chemistry! We prefer to not use MathJax in the title field, see here for details. $\endgroup$ – Martin - マーチン Jun 18 at 12:57
  • $\begingroup$ v is not the wave speed, it is the speed of the object with the given mass and de Broglie frequency. Also. there should be rather p instead of mv, unless v << c. $\endgroup$ – Poutnik Jun 18 at 13:31
  • $\begingroup$ @Poutnik How is p different from mv? Pardon me if my question is elementary, I have only just started studying quantum chem. $\endgroup$ – ijm Jun 19 at 8:13
  • $\begingroup$ I think $c$ is meant to be the wave speed here, not necessarily the speed of light. This is common usage in physics at least. $\endgroup$ – d_b Jun 21 at 0:16
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$v$ is not the wave speed, it is the speed of the object with the given mass and de Broglie frequency. Also, there should be rather $p$ instead of $mv$, unless $v \ll c$.

As $$ E = \sqrt{(mc^2)^2 + (pc)^2}$$ then $$p = \sqrt{\left({\frac {E}{c}}\right)^2 - {(mc)}^2}$$

For a massless object like a photon:

$$p = \frac Ec$$.

For an object with mass, there is also $$p = \frac {mv}{\sqrt{1-(v/c)^2}}= \gamma mv$$

See Matter wave, paragraphs De Broglie relations and Special relativity, where is clear c is the phase speed of light in vacuum:

$$\begin{align}&\lambda =\,\, \frac {h}{\gamma m_0v}\, =\, \frac {h}{m_0v}\,\,\,\, \sqrt{1 - \frac{v^2}{c^2}}\\ & f = \frac{c}{\lambda} = \frac{\gamma\,m_0vc}{h} = \frac {m_0c^2}{h} \bigg/ \sqrt{\frac{c^2}{v^2}-1} \end{align}$$

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  • $\begingroup$ "v is not the wave speed, it is the speed of the object with the given mass and de Broglie frequency." How is this different from saying that it is the speed of the electron? What is the nuance here? $\endgroup$ – ijm Jun 19 at 12:17
  • $\begingroup$ It may not be an electron, as wave properties are observable for objects of size/mass up to small molecules. $\endgroup$ – Poutnik Jun 19 at 12:28

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