Characteristic radiation - scattering When an incident electron strikes an inner shell electron, I read that the incident electron is scattered and the inner shell electron is ejected. 
What exactly does it mean for the electron to be 'scattered'? I have quite a classical picture in my head of a nucleus and electrons orbiting around the nucleus. So the electron has to 'pass' the orbitals of other electrons, 'hit' this inner shell electron and then scatters away? I have learnt quantum mechanics, but I suck at it and its all operators and we never have any qualitative explanations/what the maths actually means, so I'm stuck with my classical hat on for now...
Also, another question I just had was what if the incident electron's energy was completely transferred to the ejected electron? Surely the incident electron doesn't just 'stop'/become stationary but that's classically what would happen if, for example, a ball lost all of its kinetic energy. 
Probably loads of misconceptions here, so insight is appreciated. 
[Edit: feel free to drop plugs for good textbooks on this sort of thing too]
 A: For an introduction to QM, for anyone comfortable with calculus, I recommend Griffiths' book Introduction to Quantum Mechanics. It is relatively easy to read and relatively careful with language, and it at least occasionally supplements the math with some comments about intuition. Here, I'll attempt the opposite: I'll try to offer some intuition, occasionally supplemented with a little math (no calculus). I'll start with simpler situations that don't involve scattering.
The classical picture of a hydrogen atom as a nucleus with an electron orbiting around it is flawed because the electron does not have such a sharply-defined location. It is described by a wavefunction instead. In the lowest-energy state of a hydrogen atom, the electron's wavefunction has the form
$$
  \psi(\mathbf{x})\propto \exp\left(-\frac{|\mathbf{x}|}{a}\right)
\tag{1}
$$
where $\mathbf{x}=(x,y,z)$ are the spatial coordinates relative to the location of the nucleus and where $a$ is a constant length scale that depends on Planck's constant, the electron mass, and the charge. Equation (1) describes a so-called s orbital, which can be visualized as a spherically-symmetric cloud-like distribution that fades away with increasing distance from the nucleus. We should not think of the electron as being somewhere in that cloud, but we can think of the electron as being that cloud. The electron is a "point particle" in the sense that the possible outcomes of a measurement of its location can be arbitrarily sharply localized (in non-relativistic QM); but in general, the electron's location is not sharply defined. It is described by a wavefunction, and equation (1) is an example of this.
Equation (1) is for the state of lowest energy. States of the next-higher energy have the form
$$
  \psi(\mathbf{x})\propto 
 \exp\left(-\frac{|\mathbf{x}|}{2a}\right)\mathbf{c}\cdot\mathbf{x}
\tag{2}
$$
where $a$ is the same as before and where $\mathbf{c}$ is some constant vector. Equation (2) describes a so-called p orbital, which again can be visualized as 
a cloud-like distribution (not spherically symmetric) that fades away at sufficiently large distance from the nucleus.
Now consider something more interesting: a helium atom. A helium atom has two electrons. This is more interesting because neither electron has its own wavefunction, much less its own location. Instead, only the pair of electrons has a wavefunction. For the sake of building intuition, I'll simplify things by ignoring the fact that electrons are spin-1/2 fermions; I'll pretend than they are distinguishable spin-0 particles instead. This simplifies the math without changing the key message: only the pair of electrons has a wavefunction $\psi(\mathbf{x}_1,\mathbf{x}_2)$. Because the electrons repel each other electrostatically, the magnitude of this wavefunction goes to zero wherever the distance $|\mathbf{x}_1-\mathbf{x}_2|$ goes to zero. The wavefunction is largest wherever its arguments are such that $|\mathbf{x}_1-\mathbf{x}_2|$ is roughly equal to the size of the atom (this essentially defines the "size" of the atom), because this balances the electrons' attraction to the nucleus against their repulsion from each other. This description of the shape of the wavefunction cannot be expressed in terms of either electron individually; again, only the pair of electrons has a wavefunction.
In a more complicated atom with $N$ electrons, the message is similar: only the whole system of $N$ electrons has a wavefunction $\psi(\mathbf{x}_1,...,\mathbf{x}_N)$. The magnitude of this wavefunction depends on the relationships between its arguments $\mathbf{x}_n$. It is small wherever two of these arguments are close to each other (because electrons repel each other), and it also fades away wherever one or more of these arguments is sufficiently far from the nucleus (because electrons are held close to the nucleus).
Now, what about scattering? 
Consider a situation in which one electron is incident on an $N$-electron atom. We can take the initial wavefunction to be something like
$$
  \psi(\mathbf{x}_0,\mathbf{x}_1,...,\mathbf{x}_N)
 = f(\mathbf{x}_0)g(\mathbf{x}_1,...,\mathbf{x}_N).
\tag{3}
$$
(Again, I'm simplifying things by ignoring the fact that electrons
are indistinguishable fermions.) The function $g$ describes the $N$-electron atom. The function $f$ describes the indicent electron as some kind of imperfectly-localized cloud, and it is chosen so that when we evolve the wavefunction $\psi$ forward in time using the multi-particle Schrodinger equation, this cloud approaches the atom and eventually washes over it.
The mutli-particle Schrodinger equation includes Coulomb-repulsion terms that depend on $|\mathbf{x}_j-\mathbf{x}_k|$. In particular, there are terms that depend on $|\mathbf{x}_0-\mathbf{x}_k|$ for $k\geq 1$. As a result, the initial wavefunction evolves into an $N+1$-particle wavefunction
$$
  \psi(\mathbf{x}_0,\mathbf{x}_1,...,\mathbf{x}_N,t)
\tag{4}
$$
that can no longer be factorized as it was initially. In words, the electrons are all entangled with each other. 
At late enough times, we can end up with something like
$$
 \psi = \psi_A + \psi_B + \psi_C + \cdots
\tag{5}
$$
(where each term is a function of all $N+1$ points $\mathbf{x}_n$), where each term can be factorized in a particular way, similar to (3). One term, say $\psi_A$, might be factorizable into a function of $\mathbf{x}_0$ times a function of the other points. If this were the only term, it would describe a situation in which the incident electron emerges again, partly scattered (as a "cloud" shaped something like a spherical shell expanding outward away from the atom) and partly not scattered (as a "cloud" that just keeps moving in the original direction). Another term, say $\psi_B$, might be factorizable into something like $f(\mathbf{x}_0,\mathbf{x}_1)g(\mathbf{x}_2,...,\mathbf{x}_N)$, where the $f$ part is a kind of two-electron "cloud" propagating away from the atom and the $g$ part is what's left of the atom. If this were the only term, it would describe a situation in which one of the original atomic electrons is ejected — which could be an outer-shell electron, or an inner-shell electron, depending on which term in (5) we are considering. 
The total wavefunction (5) at time $t$ is a quantum superposition (sum) of all of these terms.
When a measurement occurs — as it  eventually does, whether or not we do it deliberately — we essentially end up with just one of these terms. Quantum theory predicts how frequently each of the possible outcomes occurs, but it cannot predict which one will occur.
I tried to simplify things by pretending that we can talk about each electron as though it had an identity of its own. In reality, electrons are indisinguishable fermions. Mathematically, this means that the wavefunction $\psi(\mathbf{x}_0,\mathbf{x}_1,...,\mathbf{x}_N)$ must change sign when any two of its arguments are exchanged with each other. (I'm still pretending that electrons have spin 0.) Because of this, when an electron is scattered or ejected from an atom, talking about which electron was scattered or ejected doesn't really make sense. All we can do is talk about how many electrons were scattered or ejected. This relates the the second part of your question: 

...what if the incident electron's energy was completely transferred to the ejected electron?

In this case, we could describe the final state by saying that one electron leaves the atom. We can't say which electron leaves the atom — whether it's the one that was incident or one that was originally bound to the atom — because electrons don't have individual identities. They can be counted, but they cannot labelled. 
I've tried to convey some intuition without using much math, which is always risky. I've also made some simplifications, which is also risky. Fixing these shortcomings would make this post a lot longer than it already is. My aim here was only to offer a little bit of intuition, without trying to make it perfect.
