Why does de Broglie wavelength lambda = c / frequency? 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} &amp;= \frac{c}{\nu} \\
\implies v &amp;= \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?
$v$ 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:

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