Application of Bohr's Theory of hydrogen to Beryllium

      The beryllium has two orbits 1s and 2s each with two electrons. For the stability of each electrons and nucleus, the two electrons are diametrically opposite in both the orbits, so that its diameters are perpendicular to each other.

                                                           Beryllium atom with 1s²2s²

      Considering 1s electrons   4e² /Kr_11s²  - e² /4Kr_11s²  = (15/4) e²/K = m v_11s²r_11s. and h/2π = m v_{11 s} r_11s The radius of the 1s orbit with two electrons becomes (4/15) a_o. and  the velocity of the electron v_11s = (15/4) e²/2hε_o. Since the electronic structures of both the orbits are same r_n = n² r₁ so that r_22s = (16/15)a_o. Since the inner orbital electrons are more tightly bound with the nucleus, its structure will remain unaltered.  Total energy of the system is sum of energies contributed by both the 1s and 2s electrons. The kinetic energy associated with the 1s electrons mv_11s² =  4e²/Kr_11s - e²/4Kr_11s = (15/4)e²/Kr_11s ^.The negative potential energy  is -8 e²/Kr_11s and the positive potential energy due to electron-electron interaction is e²/2Kr_11s.  Total energy associated with the 1s electrons is {15/4 - 8 +1/2] e²/Kr₁ = - (15/4) e²/Kr₁ = - (225/8)(e²/2Ka_o) = - 382.36  eV

     If there is no screening of nuclear charge, the radius of the 2s orbital becomes r_22s = 4 r_11s =

(16/15)a_o. Total energy associated with the 2s electrons is {15/4 - 8 +1/2] e²/Kr_22s = - (15/4) e²/Kr_{22 s} = - (225/32)[e²/2Ka_o] = - 7.03125 x 13.595 = -95.589 eV. The sum of first and second ionization energy of Beryllium atom is 9.32 +18.21 =27.53 eV.

     If we assume full screening of nuclear charge, the 2s electrons will realize only 2e⁺ . The stability of the 2s electron in its orbits requires 2e²/K - e²/4K = (7/4)e²/K = mv_22s²r_22s and h/π = m v_22s r_22s which together provide r_22s = (16/7) a_o

    The kinetic energy associated with the 2s electrons mv_22s²  =  (7/4) e²/Kr_22s .Negative potential energy  - 4 [e²/Kr_22s] _.  Positive potential energy of 2s electrons due to  the electron-electron interaction e²/2Kr_22s Total energy associated with the 2s electrons is -(7/4)[e²/Kr_22s] = -(49/32) x  [e²/2Ka_o]= -1.53125 x 13.595 = -20.8173 eV . It is little closer to the sum of the observed first and second ionization energy of the beryllium atom 9.32 + 18.21 = 27.53 eV

    The  first ionization energy of beryllium is determined from the knowledge of total energy associated with beryllium and beryllium ion Be^+.

Total energy possessed by beryllium atom = - 382.36 + - 20.82 = -403.18 eV , the observed experimental value is -398.03 eV.  In Be⁺ , the 2s electron has  kineetic energy  (1/2) m v_2s²  = e²/Kr_2s and negative potential energy - 2e² /Kr_{2s ,} Adding togethertotal energy  becomes - e²/Kr_2s

= - e²/2Ka_o = - 13.595 eV. It gives the first ionization energy of Beryllium as (20.82 - 13.60) 7.22 eV .