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                         CONCEPTS Applied                                 with matter is an “all-or-none” affair; that is, matter absorbs an
                                                                          entire photon or none of it. The relationship between frequency
                         Atomic Parts                                     (f ) and energy (E) is
                     Identify the number of protons, neutrons, and electrons                    E = hf
                                 16
                     in an atom of  8 O. Write your answer before you read the
                                                                                                                  equation 8.1
                     solution in the next paragraph.
                        The subscript to the bottom left is the atomic number.   where  h is the proportionality constant known as  Planck’s
                     Atomic number is defined as the number of protons in the            –34
                                                                          constant (6.63 × 10  J⋅s). This relationship means that higher-
                       nucleus, so this number identifies the number of protons
                                                                          frequency light, such as ultraviolet, has more energy than lower-
                     as 8. Any atom with 8 protons is an atom of oxygen,
                                                                          frequency light, such as red light.
                     which is identified with the symbol O. The superscript
                     to the top left identifies the mass number of this isotope
                     of oxygen, which is 16. The mass number is defined as
                     the sum of the number of protons and the number of   EXAMPLE 8.1
                     neutrons in the nucleus. Since you already know the
                     number of protons is 8 (from the atomic number), then   What is the energy of a photon of red light with a frequency of 4.60 ×
                                                                            14
                     the number of  neutrons is 16 minus 8, or 8 neutrons.   10  Hz?
                     Since a neutral atom has the same number of electrons
                     as protons, an atom of this oxygen isotope has 8 protons,   SOLUTION
                     8 neutrons, and 8 electrons.
                        Now, can you describe how many protons, neutrons,      f  = 4.60 ×  10     Hz
                                                                                          14
                                                    17
                     and electrons are found in an atom of  8 0? Compare your             –34
                                                                               h  = 6.63 ×  10     J⋅s
                     answer with a classmate’s to check.
                                                                              E  = ?
                                                                                    E = hf
                                                                                                            14 1 _
                                                                                               –34
                                                                                      = (6.63 ×  10     J⋅s)(4.60 ×  10             )
                                                                                                              s
                                                                                               –34        14     1 _
                                                                                      = (6.63 ×  10     )   (4.60 ×  10     )   J⋅s ×

                                                                                                                 s
                    8.2  THE BOHR MODEL                                                        –19
                                                                                      = 3.05 ×  10     J
                   Niels Bohr was a young Danish physicist who visited
                     Rutherford’s laboratory in 1912 and became very interested
                   in questions about the solar system model of the atom. He   EXAMPLE 8.2
                     wondered what determined the size of the electron orbits   What is the energy of a photon of violet light with a frequency of
                   and the energies of the electrons. He wanted to know why   7.30 × 10  Hz? (Answer: 4.84 × 10  J)
                                                                                                    –19
                                                                                 14
                   orbiting electrons did not give off electromagnetic radia-
                   tion. Seeking  answers to questions such as these led Bohr
                   to incorporate the quantum concept of Planck and Einstein
                   with Rutherford’s model to describe the electrons in the outer   ATOMIC SPECTRA
                   part of the atom. We will briefly review this quantum concept
                   before proceeding with the development of Bohr’s model of   Planck was concerned with hot solids that emit electromag-
                   the hydrogen atom.                                     netic radiation. The nature of this radiation, called blackbody
                                                                          radiation, depends on the temperature of the source. When
                                                                          this light is passed through a prism, it is dispersed into a con-
                                                                          tinuous spectrum, with one color gradually blending into the
                   THE QUANTUM CONCEPT                                    next as in a rainbow. Today, it is understood that a continuous
                   In the year 1900, Max Planck introduced the idea that mat-  spectrum comes from solids, liquids, and dense gases because
                   ter emits and absorbs energy in discrete units that he called   the atoms interact, and all frequencies within a temperature-
                   quanta. Planck had been trying to match data from spectros-  determined range are emitted. Light from an incandescent gas,
                   copy experiments with data that could be predicted from the   on the other hand, is dispersed into a line spectrum, narrow
                   theory of electromagnetic radiation. In order to match the   lines of colors with no light between the lines (Figure 8.8). The
                   experimental findings with the theory, he had to assume that   atoms in the incandescent gas are able to emit certain char-
                   specific, discrete amounts of  energy were associated with differ-  acteristic frequencies, and each  frequency is a line of color
                   ent frequencies of radiation. In 1905, Albert Einstein extended   that represents a  definite value of energy. The line spectra are
                   the quantum  concept to light, stating that light consists of dis-  specific for a substance, and increased or decreased temper-
                   crete units of energy that are now called photons. The energy of   ature changes only the intensity of the lines of colors. Thus,
                   a photon is directly  proportional to the frequency of vibration,   hydrogen always produces the same colors of lines in the same
                   and the higher the  frequency of light, the greater the energy of   position. Helium has its own specific set of lines, as do other
                   the  individual photons. In addition, the interaction of a photon   substances. Line spectra are a kind of finger print that can

                   208     CHAPTER 8  Atoms and Periodic Properties                                                       8-6
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