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Quantum 2

The Photo-electric effect:


Einstein interpreted an experiment in which others shined light of different wavelengths onto a metal surface and found that electrons could be ejected from the surface. If one turned up the intensity of the light, more electrons were ejected, but they all had the same kinetic energy. If one increased the frequency of the light (lowered the wavelength), the electrons came off at higher velocity, therefore higher kinetic energy (since the mass of the electron is constant, higher KE means higher velocity). Einstein found that the kinetic energy of the electrons was related to frequency by the equation hν=BE+KE where ν is the frequency of light used and BE is the ionization energy (Binding Energy). It may seem obvious now, but it was not always clear how a ray of light could impart energy to a particle. In order to make sense of these results, and those of Planck, Einstein proposed that energy was transferred to the electron by a particle of light called a “photon” and that the energy of a photon was given by the equation Ephoton=hν. Therefore, the reason light was quantized is that each quantum of energy corresponds to the energy of a photon. One electron was ejected by a collision with one photon. The kinetic energy of the electron after the collision was equal to the kinetic energy of the photon after the collision minus that amount needed to remove the electron. Based on this, it was possible to calculate what the momentum of the photon had to be in order to impart energy to the electron at the measured level. Using the equations one could show that the momentum (p) of the photon was given by p=hν/c=h/λ. This suggests that a photon has mass, which it really doesn’t. It does have momentum.

The Quantum Side


This is not too hard. We'll cover it later in the week.
In fact, the math is easier.
Step 1 is to figure out the energy of each level in the Bohr atom. The amount of energy on level 1 is The answer is in joules because the ionization energy is (2.18x10
-18 J ) constant is in Joules.
For Level 2 it is and for level 3 it’s 1/32 times 2.18 E
-18 etc. Figure out the value in joules of levels 2 through 5.
Now, find the difference between the value for 2 and the value for 3 (it should be about 3.03x10
-19). Do the same for the difference between 4 and 2, then 5 and 2. In each case, this is the energy of the photon that must be emitted.
Now that you have the Energy of each photon, use that to solve for the wavelength using “h” is Planck’s constant (6.63x10-34 J*s) and c is the speed of light (3.00x108m/s). Multiply hxc (you should get 1.99x10-25) and divide that by each of the energies you got for the differences between 3 and 2 etc. You should get wavelengths that are similar to the red, teal and blue lines.

That last little bit you thought was still right…


The Bohr model has a lot of truth in it:
  1. Bohr’s idea that the electron exist only in discrete energy states is correct.
  2. He is also correct that the emission or absorption of a photon occurs when an electron transitions from one energy state to another.
  3. Where he was wrong was in that his mathematical approach did not explain the emission spectra of any other atom, and he had no real explanation as to why the electron should only be in discreet states.

So, how do we calculate the energy of an electron?


  1. de Broglie and Schröedinger decided that they could take the equations developed to describe the energy of waves of fluid, or sound, and apply them to the behavior of the electron.
  2. The result was what is known as a “wave function,” which is abbreviated Ψ (psi) and is also known as an orbital. For each electron of an atom, there is a unique form of wave equation, which describes the energy of that electron. While this in not an “orbit,” this serves the same role as the “orbit” in the Bohr model, and so is called an orbital.
  3. An orbital does not tell you much about where an electron is, or where it is going. However, there is a form called Ψ2, which predicts the probability of an electron being in any particular place.
  4. Since we have had to rely on the wave mechanical model to describe the electron, it is difficult to then solve for the position of the electron (position implies it is a particle).

Werner Heisenberg came up with the following equation

Δx X Δmv≥h/4π
where Δx is the uncertainty of the location of the electron and Δmv is the uncertainty in the momentum (momentum=mass x velocity) of the electron. h/4π (Planck's Constant times 4 pi) is a small number (5.28x10-35). But, when we are talking about an electron’s mass and velocity, the number is big enough to make a difference. The result of this is that as you know more about an electrons position, you know less about its momentum and vice versa.

SO…we now can calculate the energy of an electron of other elements at any energy state (orbital)
What is the test of this?
We look to see whether it accurately tells us the wavelength of the bright-line spectrum of other elements…it does.
We also now know why an electron only exists in discreet states: this is a characteristic of waves. Once you define certain parameters (like, where the nodes are, like in a guitar string), the distribution of the electron is constrained.

What is an orbital?


  1. The wave function describes the energy of an electron. What are the types of energy we need to take into account? a. Kinetic Energy of the electron (it’s moving and it has mass)
  2. Potential Energy attracting the electron towards the nucleus (opposites attract).
  3. Potential Energy resulting from the repulsion from other electrons.
  4. Bohr dealt pretty well with the first two. However, he never was able to deal with the third. The reason his equations work for hydrogen is that there are no other electrons to repulse.
  5. The wave function can be thought of as a “ticket” for a seat in a theater.
The analogy is a little strained, but go with me here. Suppose in this theater, all the seats come in boxes with two seats. So, your ticket has to specify which box, and then which of the two seats you have. The theater has many levels, so the first thing the ticket says in what level your seat is on. It may be level 1, 2, 3, etc. Each level has sections, section, called “s” “p” “d” “f,” which your ticket specifies next. Each successive level has room for more sections (you add one section for each new level as you move away from the stage. Level one has section s, only. Level 2 has sections s and p; 3 has s, p and d; 4 has s, p, d and f) And each section has more boxes (each section “s” has only one box in it, section p has three, section d has five and section f has 7 boxes. Finally, once you know which level, which section, and which of the boxes, you specify which of the two seats. So, try to picture this theater.
Level 1 is closest to the stage, and has only section “s” which has only 1 box. Those are the best seats, but there are only two of them.
When you go to level 2, you have section “s” which has only one box (2 seats in it) and section “p” which has three boxes, total of six seats. The next best seats are on level 3, again there is a section s (one box), a section p (three boxes) and now even a section d with 5 boxes in it. This goes on. Oh, one last thing: the ticket printer has a perverse sense of humor in terms of how he names each box and seat. The single box in any ‘s’ section is called box “0,” the three boxes in the ‘p’ section are called -1, 0 and +1; the 5 boxes in section ‘d’ are called -2, -1, 0, 1, 2 and the 7 sections in section f are -3, -2, -1, 0, 1, 2, 3. And the seats (remember, only two per box) are called either +1/2 or -1/2. So, if you get a great seat, you might be in level 1, section s, box 0 and seat +1/2. Maybe I’m in level 2, section p, box -1, seat +1/2. And, just confuse things more, section “s” is also known as section “0,” p is known as 1, d as 2 and f as 3....yeah, we needed that complication. But, remember, these are really part of equations, so they have to have some numerical value.

Quantum Numbers (without the theater metaphor)


The first or principle quantum number is known as “n,” and can have positive integer values: 1; 2; 3; 4 etc. These play the same exact role as “n” in the Bohr equation. What Bohr missed is that at each primary level, there were multiple possible states for the electron to be in, and each differs slightly in energy. 2. Second quantum number is the angular momentum number and is symbolized as l . It loosely can be called the “shape” of the orbital. It depends somewhat on the principle quantum number. The values are whole numbers from 0 to (n –1). So, if principle quantum number is 1, the only value for the second is 0 (n-1). However, on the second quantum level, n=2, so l can be 0 or 1. If n=3, l can be 0; 1or 2. Oh, and just to be confusing, there are names given to these values. l=0 is known as an “s” (spherical) orbital. l=1 is a “p” (petal) orbital, l=2 is a “d”(I don’t know why) orbital and “3” is “f”. If n=1, then we can only have an s orbital. If n=2, we can have an “s” and a “p.” On n=3, we have s, p and d. Level 4 has s, p, d, and f. 3. The next quantum number is known as ml, or the “magnetic” quantum number. The values for ml, go from – l, 0, + l. That is, an “s” orbital (l=0), there is only one magnetic state. But, if we have a “p” orbital, (l=1) we can have –1, 0, and +1 as possible values. This number essentially tells you the orientation of the orbital in space. A sphere can only have one orientation. But the “petals” can be either in the x, y or z plane. That gives you three orientations, called -1, 0 and 1. For d and f, the orientations are a bit harder to explain, but there are 5 and 7 orientations respectively. 4. Finally, each ml can have two electrons in it, in what is known as a “spin pair.” This is called the “spin” quantum number and for any ml there are 2 and only two possible states, called either + and or +1/2 and –1/2. 5. So, if n=1, there is only one possible orbital, the S (l=0) state, and it can hold two electrons (or has two sites that could be occupied). If n=2, I have an “s” (l=0) and a “p” (l=1). The p orbitals comprise 3 three possible ml values, each of which has 2 possible “slots” for electrons(spin quantum number). Thus, the 3 p orbitals combined can have 6 electrons in them. Here’s the fun part: if you do the math, it turns out that this set of numbers explains the shape of the periodic table.??? There is only room for two electrons on level 1 (in the single “s” orbital). On period 1, you have only two elements with 1 and 2 electrons (H and He). On level two (period 2), you can have an s (2 electrons) and 3 p orbitals (6 electrons total). That’s 8 elements, which is how many you have. They even come in a pair ( s orbital) on the left and a group of 6 on the right (p orbitals). For a reason I’ll explain soon, you have “d” orbitals on level three...but they don’t fill until after the level 4 s electrons, in period 4. So, level 3 has the same pattern as level 2 (2 on the left, 6 on the right for 8 total). Level 4 starts out with its own “s” orbital, then goes to fill the level 3 d orbitals. That means that these electrons are added inside the level 4s electrons. There are 5 “boxes” for a total of 10 electrons...one for each of the 10 elements in the transition metals on a given period. The same pattern repeats with level 5 being the first time you get electrons in the f orbitals. These are actually the 4 f orbitals. There are 7 orbitals (boxes) for a total of 14 electrons as found in those odd rows always at the bottom (they actually belong up in the regular table). Cool stuff...well to me anyway.
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