THE NATURE OF THE ATOM

Transcription

1 CHAPTER 30 THE NATURE OF THE ATOM CONCEPTUAL QUESTIONS 1. REASONING AND SOLUTION A tube is filled with atomic hydrogen at room temperature. Electromagnetic radiation with a continuous spectrum of wavelengths, including those in the Lyman, Balmer, and Paschen series, enters one end of the tube and leaves the other end. The exiting radiation is found to contain absorption lines. At room temperature, most of the atoms of atomic hydrogen contain electrons that are in the ground state (n = 1) energy level. Since the radiation contains a continuous spectrum of wavelengths, it contains photons with a wide range of energies ( E = hf = hc / λ). In particular, it will contain photons with energies that are equal to the energy difference between the atomic states in the Lyman series. When the radiation is incident on the atoms in the tube, these photons are absorbed by the electrons. When a photon, whose energy is equal to the energy difference for a transition in the Lyman series, is absorbed by a ground state electron, that electron will make a transition to a higher energy level. Every photon of this energy will be absorbed by a ground state electron and cause a transition. The wavelength of radiation that corresponds to that particular photon energy will, therefore, be removed from the radiation. When the radiation is analyzed, the wavelengths that correspond to transitions in the Lyman series will be absent. Since most of the atoms in the tube are in the ground state (n = 1), the electron populations in the n = 2 and n = 3 states are extremely small. Therefore, any absorption lines resulting from Balmer or Paschen transitions will be extremely weak. When the radiation is analyzed, the only predominant absorption lines in the exiting radiation will correspond to wavelengths in the Lyman series. 2. REASONING AND SOLUTION Refer to the situation described in Question 1. Suppose the electrons in the atoms are mostly in excited states. Most of the electrons are in states with n > 2; therefore, Balmer and Paschen series transitions will occur, and the absorption lines in the exiting radiation will correspond to wavelengths in the Balmer and Paschen series. Since there are relatively few electrons in the ground state, only a relatively few number of photons that correspond to wavelengths in the Lyman series will be absorbed. Most of the "Lyman photons" will remain in the radiation; therefore, the exiting radiation will not contain absorption lines that correspond to wavelengths in the Lyman series. Although the absorption lines that correspond to transitions in the Lyman series are not present, there will be more absorption lines in the exiting radiation compared to the situation when the electrons are in the ground state, because absorption lines corresponding to both the Balmer and Paschen series will be present. 3. REASONING AND SOLUTION According to Equation 30.13, the energy E n of the nth atomic state in a Bohr atom is given by E n = ( 13.6 ev )Z 2 / n 2, where n = 1, 2, 3,..., and Z is equal to the number of protons in the nucleus. The energy that must be absorbed by the electron to cause an upward transition from the initial state n i to the final state n f is E fi = E f E i. Using Equation 30.13, we find that the required energy is E fi = ( 13.6 ev )Z 2 1/ n 2 2 ( f 1 / n i ), with n i, n f = 1, 2, 3,... and n f > n i. The energy required to ionize the atom from when the outermost electron

2 Chapter 30 Conceptual Questions 175 is in the state n i can be found by letting the value of n f approach infinity. The ionization energy is then E = ( 13.6 ev )Z 2 2 ( 1 / 1 / n i )= ( 13.6 ev )Z 2 2 ( 0 1 / n i )= ( 13.6 ev )Z 2 2 / n i From this expression, we see that the ionization energy is inversely proportional to the square of the principal quantum number n i of the initial state of the electron; thus, less energy is required to remove the outermost electron in an atom when the atom is in an excited state. Therefore, when the atom is in an excited state, it is more easily ionized than when it is in the ground state. 4. REASONING AND SOLUTION In the Bohr model for the hydrogen atom, the closer the electron is to the nucleus, the smaller is the total energy of the atom. This is not true in the quantum mechanical picture of the hydrogen atom. In the Bohr model, the nth orbit is a circle of definite radius r n ; every time that the position of the electron in this orbit is measured, the electron is found exactly a distance r n from the nucleus. In contrast, according to the quantum mechanical picture, when an atom is in the nth state, the position of the electron is uncertain. Even though the atomic state is well defined by the principal quantum number n, the location of the electron is not definite. When the atom is in the nth state, the electron can sometimes be found close to the nucleus, while, at other times, it can be found far from the nucleus, or at some intermediate position. In the absence of any external magnetic fields, the energy of the state is determined by the value of n, and for a given n, the position of the electron is uncertain. Therefore, it is not correct to say that in the quantum mechanical picture of the hydrogen atom, the closer the electron is to the nucleus, the smaller is the total energy of the atom. 5. REASONING AND SOLUTION a. Consider two different hydrogen atoms. The electron in each atom is in an excited state. The Bohr model uses the same quantum number n to specify both the energy and the orbital angular momentum. According to the Bohr model, if an atom is in its nth state, the corresponding energy is E n = (13.6 ev ) Z 2 / n 2, where n = 1, 2, 3,..., and Z is equal to the number of protons in the nucleus (Equation 30.13). The integer n that specifies the energy also specifies the orbital angular momentum according to L n = nh /(2π), where h is Planck's constant (Equation 30.8). Therefore, according to the Bohr model, it is not possible for the electrons to have different energies, but the same orbital angular momentum. b. The quantum mechanical model uses two different quantum numbers to specify the energy and the orbital angular momentum of an atomic electron. In the absence of magnetic fields, the energy is determined by the principal quantum number n, while the orbital angular momentum is determined by the orbital quantum number l. According to quantum mechanics, if an atom is in its nth state with orbital quantum number l, the energy is identical to that of the Bohr model, and the magnitude of the orbital angular momentum is [ h / ( 2 π )] l( l + 1), where l = 0, 1, 2,..., (n 1). Since l = 0, 1, 2,..., (n 1), different values of l are compatible with the same value of n. For example, when n = 3, the possible values of l are 0, 1, 2, and when n = 4, the possible values of l are 0, 1, 2, 3. Thus, the

3 176 THE NATURE OF THE ATOM electron in one of the atoms could have n = 3, l = 2, while the electron in the other atom could have n = 4, l = 2. Therefore, according to quantum mechanics, it is possible for the electrons to have different energies but have the same orbital angular momentum. 6. REASONING AND SOLUTION In the quantum mechanical picture of the hydrogen atom, the orbital angular momentum of the electron may be zero in any of the possible energy states. This is true because, if an atom is in its nth state with orbital quantum number l, the magnitude of the orbital angular momentum is given by [ h / ( 2 π )] l( l + 1), where l = 0, 1, 2,..., (n 1); therefore, for any value of n, the value of l may be zero. However, when the hydrogenic electron is in the ground state, n = 1, and the only possible value for the orbital quantum number is l = 0. Therefore, when the electron is in the ground state, its orbital angular momentum must be zero. 7. REASONING AND SOLUTION a. According to the Pauli exclusion principle, no two electrons in an atom can have the same set of values for the four quantum numbers n, l, m l, and m s. For a given value of the orbital quantum number l, the magnetic quantum number m l, can have 2l + 1 possible values. Each of these values can be combined with two possible values for the spin quantum number m s. Therefore, the maximum number of electrons that the lth subshell can hold is 2(2l + 1). All of the electrons in a 5g subshell have n = 5, and l = 4. Thus, the maximum number of electrons that can be contained in a 5g subshell is 2(2l + 1) = 2[2(4) + 1] = 18. Therefore, a 5g subshell can contain at most 18 electrons and cannot contain 22 electrons. b. Since 17 is less than 18, a 5g subshell could contain 17 electrons. 8. REASONING AND SOLUTION An electronic configuration for manganese (Z = 25) is written as 1s 2 2s 2 2p 6 3s 2 3p 6 3d 4 4s 2 4p 1. According to Figure 30.17, the ground state electronic configuration for manganese is written as 1s 2 2s 2 2p 6 3s 2 3p 6 3d 5 4s 2. Clearly, the electronic configuration given in the question corresponds to an excited state in which the last electron in the 3d subshell has been excited to the 4p subshell. 9. REASONING AND SOLUTION Characteristic X-rays are produced when an electron with enough energy strikes the target atom and knocks one of the K-shell (n = 1 ) electrons entirely out of the atom. An electron in one of the outer shells (n 2 ) then drops into the K-shell and emits an X-ray photon in the process. The ground state of hydrogen contains one electron in the n = 1 state, whereas, the ground state of helium contains two electrons in the n = 1 state. In both atoms, the ground state configuration contains only K-shell electrons. Even if one of the K-shell electrons were knocked out by an impinging electron, there are no electrons in higher levels to fall into the K-shell vacancy. Therefore, we would not expect hydrogen and helium atoms in their ground state to emit characteristic X-rays.

4 Chapter 30 Conceptual Questions REASONING AND SOLUTION The characteristic lines in the X-ray spectrum depend on the target material. They occur at the wavelength value that corresponds to the wavelength of the X-ray photon that was emitted when an electron from the outer shell of the atom in the target material drops down to the K-shell. Since all identical atoms of the same element contain the same energy levels, the K α and K β lines will occur at the same wavelength values for all X-ray targets made of those atoms, regardless of the voltage applied across the X-ray tube. The cutoff wavelength λ 0, however, depends on the voltage V across the X-ray tube. According to Equation 30.17, λ 0 = ( hc ) /( ev ) ; the cutoff wavelength is inversely proportional to the voltage across the tube. Therefore, when a smaller voltage is used to operate the tube, the cutoff wavelength increases. We can conclude, therefore, that when the tube is operated at a smaller voltage, the characteristic lines should occur at the same wavelengths as they did at a higher voltage, and the cutoff wavelength should increase (i.e., shift to the right on the wavelength axis). The drawing in the text, shows the X- ray spectra produced by an X-ray tube when the tube is operated at two different voltages. The features of the drawing are consistent with the conclusions reached above. 11. REASONING AND SOLUTION Bremsstrahlung X-rays are produced when the electrons decelerate upon hitting the target. In the production of X-rays, it is possible to create Bremsstrahlung X-rays without producing the characteristic X-rays. The wavelengths of the characteristic lines depend on the target material; the cutoff wavelength, however, is independent of the target material and depends only on the energy of the impinging electrons. In order to produce Bremsstrahlung X- rays without producing the characteristic X-rays, the potential difference V applied to the tube must be chosen so that the cutoff wavelength is greater than the wavelengths of the characteristic X-rays for the target material. Therefore, using Equation 30.17, the value of V must be chosen so that V = ( hc ) /( eλ 0 ), where λ 0 > λ K α. 12. REASONING AND SOLUTION The short side of X-ray spectra ends abruptly at a cutoff wavelength λ 0. This cutoff wavelength depends only on the energy of the impinging electrons. Regardless of the composition of the target, the incident electron cannot give up more than all of its kinetic energy when it is decelerated by the target. The kinetic energy that the electron has acquired upon reaching the target is given by ev (see Section 19.2), where V is the potential difference that is applied to the X-ray tube. Therefore, the cutoff wavelength depends on the value of V, and not on the composition of the target material used in the X-ray tube. 13. REASONING AND SOLUTION A laser beam focused to a small spot can cut through a piece of metal. Light rays leaving the laser are nearly parallel, so the light can be focused to a very small spot. The intensity is the electromagnetic energy per unit time per unit area (see Equation 24.4). Therefore, when the beam is focused to a spot with a very small area, the intensity delivered by the beam is very large. For example, a pulsed ruby laser with a peak power of 10 8 W can be focused to a peak intensity of W/cm 2. This intensity is large enough to ionize air or burn a hole in a piece of metal.

5 178 THE NATURE OF THE ATOM 14. REASONING AND SOLUTION The energy of a photon of light of frequency f is given by Equation 29.2: E = hf. Since c = λf for light, the energy of a photon can be written in terms of the wavelength λ as E = hc / λ. According to Table 26.2, the wavelength of green light is smaller than the wavelength of red light; therefore, the laser that produces green light emits photons that are more energetic than those emitted by the helium/neon laser.