30.11

Concepts & Calculations

The Bohr model of the hydrogen atom introduces a number of important features that characterize the quantum picture of atomic structure. Among them are the concepts of quantized energy levels and the photon emission that occurs when an electron makes a transition from a higher to a lower energy state. Example 12 deals with these ideas.

| Concepts & Calculations Example 12 | | | Hydrogen Atom Energy Levels

A hydrogen atom (Z = 1) is in the third excited state. It makes a transition to a different state, and a photon is either absorbed or emitted. Determine the quantum number nf of the final state and the energy of the photon when the photon is (a) emitted with the shortest possible wavelength, (b) emitted with the longest possible wavelength, and (c) absorbed with the longest possible wavelength.
Concept Questions and Answers What is the quantum number of the third excited state?

> Answer The lowest-energy state is the ground state, and its quantum number is n = 1. The first excited state corresponds to n = 2, the second excited state to n = 3, and the third excited state to n = 4.

When an atom emits a photon, is the final quantum number nf of the atom greater than or less than the initial quantum number ni?

> Answer Since the photon carries energy away, the final energy of the atom is less than its initial energy. Lower energies correspond to lower quantum numbers (see Figure 30-9). Therefore, the final quantum number is less than the initial quantum number.

When an atom absorbs a photon, is the final quantum number nf of the atom greater than or less than the initial quantum number ni?

> Answer When an atom absorbs a photon, the atom gains the energy of the photon, so the final energy of the atom is greater than its initial energy. Greater energies correspond to higher quantum numbers, so the final quantum number is greater than the initial quantum number.

How is the wavelength of a photon related to its energy?

> Answer The energy E of a photon is given by Equation 29.2 as , where h is Planck’s constant and f is the photon’s frequency. But the frequency and wavelength are related by , according to Equation 16.1, where c is the speed of light. Combining these relations gives , or . Thus, we see that the wavelength is inversely proportional to the energy.

Solution

a. | When a photon is emitted with the shortest possible wavelength, it has the largest possible energy, since the wavelength is inversely proportional to the energy. The largest possible energy arises when the electron jumps from the initial state (ni = 4) to the ground state (nf = 1), as shown by transition A in Figure 30-36. Therefore, the quantum number of the final state is . The energy E of the photon is the difference between the energies of the two states, so . The energy of the nth state is given by Equation 30.13 as , so the energy of the photon is

Figure 30-35 A hologram of a galleon spaceship approaching earth. Two views of the same hologram are shown: (a) looking at the hologram perpendicularly and (b) looking at an angle to the right of the perpendicular. (©Holographics North, Inc., and Sprint. Reproduced with permission of Sprint Corporation)

Figure 30-36 A photon is emitted when an electron in the n = 4 state jumps to either the n = 1 or the n = 3 state. An electron makes a transition from the n = 4 state to the n = 5 state when a photon of the proper energy is absorbed.

b. | When a photon is emitted with the longest possible wavelength, it has the smallest possible energy. The smallest possible energy arises when the electron jumps from the initial state (ni = 4) to the next lower state (nf = 3), as shown by transition B in Figure 30-36. Therefore, the quantum number of the final state is . The energy E of the photon is the difference between the energies of the two states, so :

c. | When a photon is absorbed by the hydrogen atom, the electron jumps to a higher energy state. The photon has the longest possible wavelength when its energy is the smallest. The smallest possible energy change in the hydrogen atom arises when the electron jumps from the initial state (ni = 4) to the next higher state (nf = 5), as shown by transition C in the drawing. Therefore, the quantum number of the final state is . The energy E of the photon is the difference between the energies of the two states, so :

The next example reviews the physics of how a X-ray is produced, how its energy is related to the ionization energies of the target atom, and how to determine the minimum voltage needed to produce it in an X-ray tube.

| Concepts & Calculations Example 13 | | |

(The Production of X-Rays) The K-shell and L-shell ionization energies of a metal are 8979 eV and 951 eV, respectively. (a) Assuming that there is a vacancy in the L shell, what must be the minimum voltage across an X-ray tube with a target made from this metal to produce X-ray photons? (b) Determine the wavelength of a photon.
Concept Questions and Answers How is the photon produced and how much energy does it have?

> Answer A photon is produced when an electron in a metal atom jumps from the higher-energy n = 2 state to the lower-energy n = 1 state (see transition A in Figure 30-37). The energy E of the photon is the difference between the energies of these two states: .

Figure 30-37 A X-ray photon is emitted when an electron jumps from the n = 2 state to the n = 1 state (transition A). The K-shell ionization energy is the energy required to raise an electron from the n = 1 to the state, where it has zero total energy (transition B). Similarly, the L-shell ionization energy is the energy needed for the transition from n = 2 to (transition C). For clarity, the energy levels are not drawn to scale.

What must be the minimum voltage across the X-ray tube to produce a photon?

> Answer It takes energy to produce a photon, and this energy comes from the electrons striking the target in the X-ray tube. According to Equation 19.3, the energy possessed by each electron when it arrives at the target is eV, where e is the magnitude of the electron’s charge and V is the potential difference across the tube. To cause the target to emit a X-ray photon, the impinging electron must have enough energy to create a vacancy in the K shell. The vacancy is created because the impinging electron provides the energy to move a K-shell electron into a higher energy level or to remove it entirely from the atom. In the present case, with a vacancy in the L shell, the impinging electron can create the K-shell vacancy if it has the energy needed to elevate a K-shell electron to the L shell, the energy needed being (see Figure 30-37). This, then, is the minimum energy that the impinging electron must have. Correspondingly, the minimum voltage across the tube must be such that , or .

What is meant by the phrases “K-shell ionization energy” and “L-shell ionization energy”?

> Answer The K-shell ionization energy is the energy needed to remove an electron in the K shell (ni = 1) completely from the atom (see transition B in Figure 30-37). The removed electron is assumed to have no kinetic energy and is infinitely far away so that it has no electric potential energy. Thus, the total energy of the removed electron is zero. Similarly, the L-shell ionization energy is the energy needed to remove an electron in the L shell (ni = 2) completely from the atom (see transition C in Figure 30-37).

What does the difference between the K-shell and L-shell ionization energies represent?

> Answer As transitions B and C in Figure 30-37 suggest, the difference between the two ionization energies is equal to the difference between the total energy E2 of the L shell and the total energy E1 of the K shell.

Solution

a. | The minimum voltage across the X-ray tube is . But the difference in energies between the n = 2 and n = 1 states is equal to the difference in energies between the K-shell and L-shell ionization energies, or . Thus, the minimum voltage is

b. | The wavelength of the photon is , where f is its frequency. The frequency is related to the energy of the photon by Equation 29.2, , where h is Planck’s constant. Combining these relations gives

At the end of the problem set for this chapter, you will find homework problems that contain both conceptual and quantitative parts. These problems are grouped under the heading Concepts & Calculations.

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