CHAPTER 30THE NATURE OF THE ATOMCONCEPTUAL 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 radiationis found to contain absorption lines.At room temperature, most of the atoms of atomic hydrogen contain electrons that are in theground state (n = 1) energy level. Since the radiation contains a continuous spectrum ofwavelengths, it contains photons with a wide range of energies (E=hf=hc /λ). In particular, it willcontain photons with energies that are equal to the energy difference between the atomic states in theLyman series. When the radiation is incident on the atoms in the tube, these photons are absorbed bythe electrons. When a photon, whose energy is equal to the energy difference for a transition in theLyman series, is absorbed by a ground state electron, that electron will make a transition to a higherenergy level. Every photon of this energy will be absorbed by a ground state electron and cause atransition. 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 thatcorrespond to transitions in the Lyman series will be absent. Since most of the atoms in the tube arein 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 willcorrespond to wavelengths in the Lyman series._____________________________________________________________________________________________2. REASONING AND SOLUTION Refer to the situation described in Question 1. Suppose theelectrons 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 exitingradiation will correspond to wavelengths in the Balmer and Paschen series. Since there are relativelyfew electrons in the ground state, only a relatively few number of photons that correspond towavelengths in the Lyman series will be absorbed. Most of the "Lyman photons" will remain in theradiation; therefore, the exiting radiation will not contain absorption lines that correspond towavelengths in the Lyman series. Although the absorption lines that correspond to transitions in theLyman series are not present, there will be more absorption lines in the exiting radiation compared tothe situation when the electrons are in the ground state, because absorption lines corresponding toboth the Balmer and Paschen series will be present._____________________________________________________________________________________________3. REASONING AND SOLUTION According to Equation 30.13, the energy En of the nth atomicstate in a Bohr atom is given by En=– 13.6 eV()Z2/ n2, where n = 1, 2, 3, . . ., and Z is equal to thenumber of protons in the nucleus. The energy that must be absorbed by the electron to cause anupward transition from the initial state ni to the final state nf is ∆Efi=Ef– Ei. UsingEquation 30.13, we find that the required energy is ∆Efi=– 13.6 eV( )Z21/ nf2–1 / ni2(), with ni, nf= 1, 2, 3, . . . and nf>ni. The energy required to ionize the atom from when the outermost electronChapter 30 Conceptual Questions 175is in the state ni can be found by letting the value of nf approach infinity. The ionization energy isthen ∆E∞=– 13.6 eV( )Z21 /∞– 1 / ni2()=– 13.6 eV( )Z20 – 1 / ni2()=13.6 eV( )Z2/ ni2From this expression, we see that the ionization energy is inversely proportional to the square of theprincipal quantum number ni of the initial state of the electron; thus, less energy is required to removethe outermost electron in an atom when the atom is in an excited state. Therefore, when the atom isin 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 electronis to the nucleus, the smaller is the total energy of the atom. This is not true in the quantummechanical picture of the hydrogen atom.In the Bohr model, the nth orbit is a circle of definite radius rn; every time that the position of theelectron in this orbit is measured, the electron is found exactly a distance rn from the nucleus. Incontrast, according to the quantum mechanical picture, when an atom is in the nth state, the positionof the electron is uncertain. Even though the atomic state is well defined by the principal quantumnumber n, the location of the electron is not definite. When the atom is in the nth state, the electroncan sometimes be found close to the nucleus, while, at other times, it can be found far from thenucleus, or at some intermediate position. In the absence of any external magnetic fields, the energyof 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, thecloser the electron is to the nucleus, the smaller is the total energy of the atom._____________________________________________________________________________________________5. REASONING AND SOLUTIONa. Consider two different hydrogen atoms. The electron in each atom is in an excited state. TheBohr model uses the same quantum number n to specify both the energy and the orbital angularmomentum. According to the Bohr model, if an atom is in its nth state, the corresponding energy isEn= −(13.6 eV )Z2/ n2, where n = 1, 2, 3, . . ., and Z is equal to the number of protons in thenucleus (Equation 30.13). The integer n that specifies the energy also specifies the orbital angularmomentum according to Ln=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 thesame orbital angular