Is the electron considered to be at rest within an atom? According to older models of the atom, the electron moved around the nucleus at great speeds.  If this were so, would this not mean under the laws of special relativity that it would cause significant accumulation of its mass?  I am assuming under current theories that the electron is considered to be at rest WRT to the nucleus and therefore exempt from any such effects.
 A: Let's consider Hydrogen for concreteness. Then, when the Hydrogen atom is in its ground state, the electron's probability is mostly bound within a distance $a_0$ of the nucleus of the Hydrogen atom, where
\begin{equation}
a_0 = \frac{\hbar}{\alpha m_e c} = 5.3 \times 10^{-11}\ {\rm m}
\end{equation}
where $\hbar$ is Planck's constant, $c$ is the speed of light, $\alpha$ is the fine structure constant, and $m_e$ is the mass of the electron. The length scale $a_0$ is the Bohr radius.
Because the electron is localized to a region of order $\Delta x \approx a_0$, because of the uncertainty principle $\Delta p \Delta x >\hbar/2$, there must be a corresponding uncertainty in momentum
\begin{equation}
\Delta p > \frac{\hbar}{2a_0}
\end{equation}
In fact, in a more careful calculation of the variance of the electron's momentum in Hydrogen's ground state, it turns out that the electron approximately saturates this bound.
From this information, we can estimate the "velocity" of the electron (I put scare quotes around velocity for reasons we'll get to in a minute)
\begin{equation}
v \approx \frac{\Delta p}{m_e} \approx \frac{\hbar}{2 a_0 m_e} = 1.1 \times 10^6\ {\rm m\ s^{-1}} = 3.6\times 10^{-3} c
\end{equation}
So in fact, the electron is not moving relativistically in a Hydrogen atom. There are some relativistic effects that are visible in the fine structure of Hydrogen, but these are small effects.
Now, let's get back to the scare quotes. It's important to recognize that a quantum particle does not have a trajectory $x(t)$ like we have in classical mechanics, so the concept of velocity as the time derivative of a trajectory simply doesn't exist in quantum mechanics. Being able to define both the particle's position and velocity precisely violates the Heisenberg uncertainty principle.
However, we can define the (magnitude of the) "velocity" in analogy to classical mechanics as "a measurement of the (magnitude of the) particle's momentum, divided by its mass." In this sense, the electron is not at rest, since a typical measurement of velocity would give a value of order $3.6\times 10^{-3} c$, rather than $0$. In fact, the uncertainty principle tells us that since we have localized the position of the electron within the Hydrogen atom, there must be some spread in momenta (or velocities) that include a non-zero value. The only way to have a quantum particle perfectly at rest -- meaning that a measurement of its momentum would always yield zero -- would be to have a completely de-localized particle. (Actually, that's the only way to have any definite value of momentum, not just zero).
A: The reason the planetary model of the atom would not work was that in classical electrodynamics the rotating electrons would radiate away their kinetic energy and fall on the nucleus , so no atoms could exist. Bohr in his successful model that explained the Balmer series of the hydrogen spectrum, imposed by hand angular momentum quantization . (second page).
This model is outdated by the theory of quantum mechanics which solved the hydrogen atom using its postulates. In quantum mechanics there are no orbits for the electrons to have to decide on their velocity . There are orbitals, given by the probability of the electron to be in a specific (x,y,z,t), see here for experiment..
Thus your question

Is the electron considered to be at rest within an atom?

"At rest" has no meaning within the successful theory of quantum mechanics. The relativistic effects will be inherent in the wavefunction solution for the particular atom, and will affect the orbitals (probabilities) accordingly.
A: The motion of an object is an idea, which reaches deeply in classical physics. By describing an objects by its position and momentum we were able to predict the position of the planets, as well as do ballistic calculations. However, this idea of motion is not part of the basic building blocks of quantum mechanics. In fact, in quantum mechanics it is not meaningful to ask, what an object does, if we are not observing it. Quantum mechanics does not tell us what the object does, but only tells us the probability of observing a given measurement result. Therefore, the wave function is all we have as a description of the electron, which is part of an hydrogen atom -- there is no mechanical model behind quantum mechanics, which tells us why the electron behaves this way.
Think of the two-slit experiment: There is no apparatus, which allows us to tell through which slit the object (=light, electron, ...) went without destroying the interference pattern on the screen. If we which to maintain the interference pattern, we are unable to determine what the object does between its emission and our observation on the screen. This idea can be transferred to your question: If we were able to continuously observe the electron inside the hydrogen atom, we would alter its wave function. Therefore, we should expect that the obtained result differs from the one we obtain, if we are not continuously observing the electron. Does this make sense to you?
