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56                            The hydrogen atom and the periodic table

     The negative sign of the energy  which substituted in eqn (4.15) gives
     means only that the energy of this
                                                                  me 4
     state is below our chosen zero                          E =–     .                     (4.18)
                                                                   2 2
     point. [By writing the Coulomb                              8  h
                                                                   0
     potential in the form of eqn (4.2)
                                   Thus, the wave function assumed in eqn (4.13) is a solution of the differential
     we tacitly took the potential en-
                                   equation (4.12), provided that c 0 takes the value prescribed by eqn (4.17). Once
     ergy as zero when the electron is
                                   we have obtained the value of c 0 , the energy is determined as well. It can take
     at infinity.]
                                   only one single value satisfying eqn (4.18).
                                     Let us work out now the energy obtained above numerically. Putting in the
                                   constants, we get
                                                                         )
                                                    (9.1 × 10 –31 )(1.6 × 10 –19 4  kg C 4
                                             E =–
                                                              ) (6.63 × 10
                                                  8(8.85 × 10 –12 2     –34 2  2  –2 2 2
                                                                          ) F m J s
                                               = –2.18 × 10 –18  J.                         (4.19)
                                   Expressed in joules, this number is rather small. Since in most of the sub-
                                   sequent investigations this is the order of energy we shall be concerned with,
                                   and since there is a strong human temptation to use numbers only between 0.01
                                   and 100, we abandon with regret the SI unit of energy and use instead the elec-
                                   tron volt, which is the energy of an electron when accelerated to 1 volt. Since

                                                         1eV = 1.6 × 10 –19  J,             (4.20)
     From experimental studies of the  the above energy in the new unit comes to the more reasonable-looking
     spectrum of hydrogen it was   numerical value
     known, well before the develop-
                                                            E = –13.6 eV.                   (4.21)
     ment of quantum mechanics, that
     the lowest energy level of hydro-  What can we say about the electron’s position? As we have discussed many
     gen must be –13.6 eV, and it was  times before, the probability that an electron can be found in an elementary
     a great success of Schrödinger’s  volume (at the point r, θ, φ) is proportional to |ψ| —in the present case it is
                                                                           2
     theory that the same figure could  proportional to exp(–2c 0 r). The highest probability is at the origin, and it de-
     be deduced from a respectable-  creases exponentially to zero as r tends to infinity. We could, however, ask a
     looking differential equation.  slightly different question: what is the probability that the electron can be found
                                   in the spherical shell between r and r +dr? Then, the probability distribution
                                   is proportional to
                                                            2
                                                                   2 –2c 0 r
                                                               2
                                                           r |ψ| = r e  ,                   (4.22)
                                   which has now a maximum, as can be seen in Fig. 4.2. The numerical value of
                                   the maximum can be determined by differentiating eqn (4.22)
                                                    d  2 –2c 0 r   –2c 0 r    2
                                                     (r e   )=0=e      (2r –2c 0 r ),       (4.23)
                                                   dr
                                   whence
                                                                2
                                                         1   4π    0
                                                     r =   =   2    = 0.0528 nm.            (4.24)
                                                        c 0   e m
                                   This radius was again known in pre-quantum-mechanical times and was called
                                   the radius of the first Bohr orbit, where electrons can orbit without radiating.
     Niels Bohr, Nobel Prize, 1922.
                                   Thus, in quantum theory, the Bohr orbit appears as the most probable position
                                   of the electron.
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