Reading Quiz

Question 1:

Answer:

1. When more than one state has the same energy. The ground state has degenracy of 1 (or 2 if you count spin). The second level has deceneracy of 4 (or 8 if you count spin.
2. An energy level is degenerate when that energy level corresponds to more than one independent state. The degeneracy of the ground state of hydrogen is zero. The degeneracy of the second excited state is 4.
3. An energy level is degenerate when there is more than one independent state that produces the same energy level. The degeneracy of the ground state of hydrogen is 1 (that is, only one independent states have that energy) For the second excited state, there are 9 independent states.
4. Degenerate means that an energy level n has more that one independent state (n,l,m). Hydrogen's ground state is not degenerate; it's 2nd excited state has degeneracy 9.
5. An energy level is degenerate when it corresponds to more than one independent state. The degeneracy of the ground state of Hydrogen is 1. The degeneracy of the second excited state of hydrogen is 4.
6. pdVr is small enough to be ignored, and dN is basically 0, so we can ignore that as well
7. An energy level is degenerate when it correponds to more than one independent state. The ground state of hydrogen is not degenerate since it corresponds to only one state. The second excited state of hydrogen has a degeneracy of 9, since it corresponds to nine different states at that energy.
8. Degenerate energy levels are those which correspond to several states. There is only one state with the ground state energy level, so its degenracy is 1. The second excited state has a degeneracy of 9
9. When an energy level has more than one way of arranging itself such that there are multiple ways to achieve it. Hydrogen has only one ground state with energy -13.6 the degenerecy is 4.
10. A degenerate energy state is one that is debauched. Or, it is when an energy level corresponds to more than one independent state. The ground state of hydrodgen is not degenerate. The second excited state has a degeneracy of four.

Question 2:

Answer:

1. dV is zero because any change in volume is negligable compared to the corresponding change in energy. dN is zero because we don't change the number of particles (especially in a single hydrogen atom!)
2. The (mu*dN) term is neglected because dN is zero for a system of a single atom. The (PdV) term is neglected because it is much smaller than dU_R, and is thus negligible.
3. the PdV term is neglected because it is much smaller (in the case of the single atom example) thanthe dU term. For the mu*dN term, dN is really zero, since one atom stays as one atom.
4. PdV is much, much smaller than dU because an atom's volume doesn't change very much when changing state. dN is zero for for a single atom system.
5. The PdV is usually non-zero but much smaller than the dUR term and therefore negligible. The mu*dN term is neglected because dN is zero for a single atom system and other systems discussed in this chapter.
6. the boltzman distribution, it lets you determine the probability of certain energy states.
7. PdV is neglected because the change in volume when an atom is excited is tiny compared to the change in energy dU. mudN is neglected because dN is zero, at least for the systems we are currently considering (like a single atom).
8. PdV is ignored because it is very small compared to dU. µdN is thrown out because N is not changing.
9. PdV is usually nonzero, but much smaller than dU. dN is usually zero when the system is a single atom.
10. PdV can be thrown out because it is really really small, and udN was zero for 6.5 since the system is only dealing with a single atom.

Question 3:

Answer:

1. p(s) = 1/Z * exp(-E(s)/kT) It converts a botlzman factor of an energy state into a probability.
2. The most useful formula in all of statistical mechanics is: (script-P)(s) = (1/Z)*exp(-E(s)/k*T) This formula lets you determine the probability of finding an atom in a particular state s.
3. The "most [..] mechanics" is P(s)=(1/Z)*exp(-E(s)/(kT)). The formula lets you determine the probability of finding a state s at the corresponding energy E(s).
4. Boltzmann distribution: P(s) = 1/Z exp[-E(s)/kT] This is the probability of finding an atom in any particular state.
5. Equation 6.8, rho(s)=(1/Z) * e^(-E(s)/kT).
6. .6 trillion
7. The most useful formula in all of statistical mechanics is the Boltzmann distribution, which gives you the probability of a state, depending on a constant times a Boltzmann factor (which is an exponential depending on the energy and temperature of the state).
8. It provides a temperature-dependent probability distribution of all states with given energy, E(s)
9. 1/Z e^ (-E/kT), the probability distribution of a Boltzmann distribution.
10. (6.8) is the most useful formula, and it lets us determine the probability of a state s.

Question 4:

Answer:

1. The difference in energy is 12.1 eV so the ratio is approx 3.1 * 10 ^ -11. So there will be approx 31 in each of the 2nd excited states. Since there are 9 states this is approx 279 atoms in the 2nd exicted state for a tillion in the ground state.
2. 5600 are in the second excited state.
3. Well, the ratio of probabilities is P_second(s)/P_ground(s)=exp[-(E2-E1)/(kT)] which is about. 3.06*10^11. However, there are 9 states, so the probability due to degeneracy is 9 times that: 2.75*10^10. For a trillion (10^12) atoms in the ground state therefore, about 275 atoms are in the second excited state at 5800 K.
4. 31
5. 1400 atoms will be in this second excited state.
7. difference in energy is 12.1 eV, kt = 0.50 eV, ratio of probabilites is e^(-24.2) = 3 x 10^-11 = 30 x 10^-12 So for every trillion atoms in the ground state, there are on average 30 in any one of the second excited states, and since there are 9, that makes 270 atoms total in the second excited state.
8. difference in E is 12.1 eV. ratio of probabilities is e^-24.2= 3.09 x 10^-11 so for every trillion, 31 are in the second excited state
9. 0.00001764
10. 1400

Question 5:

Answer:

2. I have nothing at this time.
3. none
5. This reading seemed pretty easy to grasp, as long as my concepts of degeneracy and probability are clear.
6. I'm a little iffy going from equation (6.7) to (6.8), and pulling out that constant...