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424                                  ATOMIC PHYSICS                              [CHAP. 34



        SOLVED PROBLEM 34.11
              A sample of hydrogen gas is bombarded by a beam of electrons. How much energy must the electrons
              have if the first line of the Balmer spectral series, corresponding to a transition from the n = 3 state to
              the n = 2 state, is to be radiated?

                  The energy of the n = 3 state in hydrogen is
                                                E 1  −13.6eV
                                           E 3 =  =         =−1.5eV
                                                3 2    9
              The difference in energy between the ground (n = 1) state and the n = 3 state is 13.6 eV − 1.5 eV = 12.1 eV, so
              the energy needed by the bombarding electrons is 12.1 eV. The reason the energy difference between the n = 1 and
              n = 3 states is involved is that the hydrogen atoms are initially in the n = 1 state.


        QUANTUM THEORY OF THE ATOM
        The Bohr atomic model only applies to the hydrogen atom, the simplest one. The more general (and complex)
        quantum theory of the atom holds for many-electron atoms as well. Four quantum numbers are needed to describe
        the physical state of an atomic electron, in place of the single quantum number of the Bohr model. These are
        shown in Table 34.1.
            The possible energies of the electron are chiefly determined by n and only to a smaller extent by l and m l .
                                     2
        For the hydrogen atom, E n = E 1 /n in the quantum theory as in the Bohr theory.
            Every electron behaves in certain respects as though it is a spinning charged sphere. The amount of spin is
        the same for every electron, but there are two possible directions in which the angular momentum vector can
                                                            1
                                        1
        point in a magnetic field: “up” (m s =+ ) and “down” (m s =− ).
                                        2                   2
                                                 Table 34-1
                        Name       Symbol   Possible Values      Quantity Determined

                     Principal       n      1, 2, 3, ...    Electron energy
                     Orbital         l      0, 1, 2, ..., n − 1  Magnitude of angular momentum
                     Magnetic       m l     −l,..., 0,..., +l  Direction of angular momentum
                                             1
                     Spin magnetic  m s     − , + 1         Direction of electron spin
                                             2   2




        ATOMIC ORBITALS
                                                   2
        The distribution in space of the probability density ψ of an atomic electron depends on its quantum numbers
                                                        2
        n,l, and m l . As mentioned above, the larger the value of ψ in a certain place at a certain time, the greater the
                                                                            2
        likelihood of finding the electron there. The quantum theory of the atom enables ψ to be calculated for any
                                                                     2
        combination of n,l, and m l values for any atom. The region in space where ψ is appreciable for a given quantum
        state is called an orbital.

        ATOMIC STRUCTURE
        The two basic rules that govern the electron structures of many-electron atoms are as follows:
            1. An atom is stable when all its electrons are in quantum states of the lowest energy possible.
            2. Only one electron can occupy each quantum state of an atom; this is the exclusion principle. Thus each
               electron must be described by a different set of quantum numbers n,l, m l , m s .
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