Possible Lab Final Scenarios

Double/Single Slit
Diffraction Grating
LRC/RL CIRCUITS!!!!!!!
OSCILLISCOPE LAB!!!!
Lenses
Emission Spectra

review and memorize the circuit diagram for the LRC/oscilliscope/function generator lab!!

Good luck guys! especially Lauren cause you actually read these

just in case anyone cares

I'm sure you chem people know about the Aufbau Principle which basically states that electrons fill in the lower energy levels first and then move their way up from there. It is one of the most basic theories that is the basis for all of chemistry. Well, the Stanford Particle Accelerator Lab has just discovered with their X-ray picoscope (which is an actual thing I didn't make it up it means a "microscope" that uses particles with a wavelength of 10^-12) which can actually watch chemical reactions occur at the atomic level, that they in fact do not follow this principle, but randomly fill in whatever orbital is available...so in conclusion Chem sucks; Physics rocks!

Quantum Numbers cont'd

So in order to use this procedure with particular atoms we have to start with Hydrogen because we know that Hydrogen has 1 electron, which is in the n = 1 shell. So beginning with hydrogen:

Element: Hydrogen, # of electrons = 1

n = 1, l = 0, m_l = 0, m_s = +1/2 (or you could write n = 1, l = 0, m_l = 0, m_s = -1/2), but since there is only one electron you only write one of these two options.

Now as we go down and continue to add these numbers the number of electrons will coincide with each element. The next one being Helium because it has two electrons:

Element: Helium, # of electrons = 2

n = 1, l = 0, m_l = 0, m_s = +1/2

n = 1, l = 0, m_l = 0, m_s = -1/2

Then, you continue down the line:

Element: Li, # of electrons = 3

n = 1, l = 0, m_l = 0, m_s = +1/2

n = 1, l = 0, m_l = 0, m_s = -1/2

n = 2, l = 0, m_l = 0, m_s = +1/2

So basically you just list everything in order and for each electron you count them and put the atom by the # of electrons it has

So write the configurations for Oxygen

Quantum Numbers

Ok so it sounds like some of you are confused about quantum numbers so here we go:

Each set of quantum numbers, which will be a set of four numbers separated by commas, represents one electron and said electrons "position" in the atom. According to the Pauli Exclusion Principle, which you should know, no two electrons can have the same 4 quantum numbers. The quantum numbers each have rules discussing which numbers they are "allowed" to be, which are as follows:

n = 1, 2, 3, 4,...,infinity

l = 0, 1, 2, 3, ... , n - 1

m_l = -l , ... ,+l

m_s = -0.5, +0.5

Therefore if I have a certain n number like n = 3 I can have electrons with angular momentum numbers, l numbers, up to n - 1; which for us is 3 - 1 = 2. Thus, the L-numbers possible are l = 0, 1, 2 and thats it. IN EACH OF THOSE I can have a magnetic quantum number, m_l-number, going from -l to +l. Then in each of those we can have an electron spin of -1/2 or +1/2. So for n = 3 we have these possible electron positions:

n = 3, l = 0, m_l = 0, m_s = +1/2

n = 3, l = 0, m_l = 0, m_s = -1/2

notice that these two electrons have all the same quantum numbers except for the m_s number, or the electron spin number. Continuing on:

n = 3, l = 1,

now since l = 1 my m_l numbers can either be -1, 0, +1

n = 3, l = 1, m_l = -1,

now there are two more possibilities for m_s:

n = 3, l = 1, m_l = -1, m_s = -1/2

n = 3, l = 1, m_l = -1, m_s = +1/2

and when m_l = 0:

n = 3, l = 1, m_l = 0, m_s = -1/2

n = 3, l = 1, m_l = 0, m_s = +1/2

and finally when m_l = +1:

n = 3, l = 1, m_l = +1, m_s = -1/2

n = 3, l = 1, m_l = +1, m_s = +1/2

Now try and write all of the configurations for when n = 3, l = 2