If electrons behave as standing waves when they are bound to an atom then how do they carry charge? Today in my physics lesson we learnt that the best way of describing the behaviour of an electron that is bound to an atom is to treat it as a standing wave. I understand that this is the wave-particle duality however I was taught that in the wave-particle duality something either acts as a wave or a particle, not both at the same time. With this in mind I'm curious as to how electrons can still carry charge when they are behaving as a wave. I asked my teacher this question and he wasn't able to answer it so hopefully someone on here can explain it to me :) 
 A: It's likely that your mental picture of the "wave" that describes the electron is misleading you.  If you're thinking that the electron itself is spread out in the form of the wave and that it's charge is too, then you should rethink your picture.
The electron "is" or "behaves like" a wave in that it's state is described by something called a wavefunction.  What does this mean?  Well the wavefunction basically tells you the probability of measuring where the electron will be if you try to measure the position of the electron.  In this picture, the electron is still being thought of as a pointlike particle with charge, but it's location is uncertain, and this uncertainty is where the "waviness" comes from.
So, for example, let's consider the electron in the Hydrogen atom.  It's best not to think of the "wavelike" nature of the electron as being represented by a wavy thing circulating around the nucleus or something of that sort.  Instead, imagine that the wavefunction of the atom determines a sort of "cloud" with regions of different density where the electron is most likely to be found if one were to measure its position, like in the following image:

You might find this hyperphysics link illuminating in this regard.
A: According to de-Broglie, the wavelength associated with a particle of mass m, moving with velocity v is given by the relation,
$$\lambda=\frac{h}{mv}=\frac{h}{p}$$
where h is Planck's constant, v is the velocity and p(=mv) is momentum of the particles. The waves associated with material pariticles are called de Broglie waves.   


*

*The wavelength of an electron with mass $9.11*10^{-31}$kg and moving with the velocity of $10^6$ m/s is $7.28$ m as shown below:
$$\lambda=\frac{h}{mv}=\frac{6.63*10^{-34}kg m^2 s^{-1}}{(9.11*10^{-31}kg)(10^6m/s)}=7.28*10^{-10}m$$  

*Let's check the wavelength associated with a car of mass $10^{6}$ kg and moving with velocity of $9.11*10^{-31}$ m/s (almost at rest):  
$$\lambda=\frac{h}{mv}=\frac{6.63*10^{-34}kg m^2 s^{-1}}{(10^6kg)(9.11*10^{-31}m/s)}=7.28*10^{-10}m$$
This shows that a car or any material object with mass $10^{6}$kg and moving with velocity $9.11*10^{-31}$m/s (almost at rest) has the wavelength same as that of an electron. But, you are unable to see the wavelength associated with it because it is far smaller than the macroscopic car size. Still there is a wavelength associated with it but you are unable to see it. Because of negligible wavelength, car is said to show particle nature.   
Similarly, electron has same wavelength as that of car. Here its wavelength is far greater than its size (radius of $2.817*10^{-15}$m). So, wave character predominates. Still there is particle (which carries charge), but not observed significantly because of its greater wavelength with respect to it. Here, because of greater wavelength, electron is said to show wave nature.   
I hope this helped you.
