Physics for Scientists and Engineers, 6e Chapter 43 - Molecules and Solids

A gas of diatomic molecules is absorbing electromagnetic radiation over a wide range of frequencies. Molecule 1 is in the J = 0 rotation state and makes a transition to the J = 1 state. Molecule 2 is in the J = 2 state and makes a transition to the J = 3 state. The ratio of the frequency of the photon that excited molecule 2 to that of the photon that excited molecule 1 is impossible to determine

Equation 43.7 shows that the energy spacing is proportional to J, the quantum number of the higher state in the transition. Because the frequency of the absorbed photon is proportional to the energy separation of the states, the frequencies are in the same ratio as the energy separations.

A gas of diatomic molecules is absorbing electromagnetic radiation over a wide range of frequencies. Molecule 1, initially in the v = 0 vibrational state, makes a transition to the v = 1 state. Molecule 2, initially in the v = 2 state, makes a transition to the v = 3 state. The ratio of the frequency of the photon that excited molecule 2 to that of the photon that excited molecule 1 is impossible to determine

This is similar to question 2, except that the vibrational states are all separated by the same energy difference.

In the figure below, there is a gap between the two sets of peaks. This is because no pair of energy levels are separated by an energy corresponding to this frequency 2.the transition between energy levels that are separated by an energy corresponding to this frequency is not allowed 3.molecules cannot rotate and vibrate simultaneously

The J = 0 to J = 0 transition is forbidden according to the selection rules.

A spectrum like that in the figure below is taken when the HCl gas is at a much higher temperature. In this new spectrum, the absorption peak with the highest intensity is the same as the one in the figure 2.is farther from the gap than in the figure 3.is closer to the gap than in the figure

At a higher temperature, transitions from higher excited states will be more likely, so that peaks farther from the gap will be more intense. In Equation 43.15, at a higher temperature, the exponential factor does not decrease as rapidly with J if the temperature is higher.

In a piece of silver, free-electron energy levels are measured near 2 eV and near 6 eV. (The Fermi energy for silver is 5.48 eV.) Near which of these energies are the energy levels closer together? eV 2.6 eV 3.The spacing of energy levels is the same near both energies.

In Equation 43.21, the density of states increases with energy. Thus, at the higher energy of 6 eV, the states are closer together.

Consider the situation in question 6 again. Near which of these energies are there more electrons occupying quantum states? eV 2.6 eV 3.The number of electrons is the same near both energies.

Because the energy of 6 eV is higher than the Fermi energy, there are essentially no electrons in states near 6 eV.

Figure 43.29a shows a photon of frequency f = E g /h being emitted from a semiconductor. In addition to photons of this frequency, photons with no other frequencies are emitted 2.photons with lower frequencies are emitted 3.photons with higher frequencies are emitted

Photons can be emitted when electrons make a transition from anywhere in the conduction band above its bottom to anywhere in the valence band below its top. The lowest-energy transition, resulting in the lowest-frequency photon, is that shown in Figure 43.29a.