What is the mass of an electron in the sense of its wave nature? I just completed a course in Mechanics and I'm currently doing electromagnetism. I haven't rigorously started QM; or Modern Physics. I read up a few articles on the Wave-Particle duality.
So, how is the mass defined for an electron while it exhibiting its wave nature?
 A: Mass is mostly a notion which shows up in dynamical interactions, but you can see it in principle in the behavior of the wave. For instance, the dispersion relation for the quantum wave corresponding to a non-relativistic electron (the relation between angular frequency $\omega$ and wave-vector $k$) will be
$$\omega = \frac{\hbar k^2}{2m}$$
This is just a "quantized" version of $E = p^2/(2m)$ with $E = \hbar \omega$ and $p = \hbar k$. 
Another example: if you have a photon with energy $E_p = \hbar \omega_p$ radiated by the electron, its energy drops to $E' = E - E_p$ and you can compute that its wave-vector will now be
$$k' = \sqrt{\frac{2 m (\omega - \omega_p)}{\hbar}}$$ 
All these relations have $m$ figuring in them and for a different $m$ you would get completely different scalings.
For more digging, you can see e.g. the wiki of De Broglie waves.
A: The effective mass of an electron,$M$ can be found from its De Broglie wavelength, $\lambda$ which is the wavelength of the electron from its wave nature by the following equations:
$$\lambda=\frac{h}{p}=\frac{h}{Mv} $$
 Where $v$ is the speed of the electron wave- particle and $p$ is its momentum.
The by relativistic effective mass of the electron, M is given by:
$$M=\frac{M_0}{\sqrt{1-v^2/c^2}}$$
Where $M_0$ is the rest mass of electron.
So thus the effective mass in terms of De Broglie wavelength can be found from the first and second equation by:
$$M=\frac{M_0\sqrt{1-v^2/c^2}}{\lambda*v}$$
