Question

Calculate the interatomic distance for each of the following molecules when the nuclear repulsion is equal to 15 eV.

a) H2

b) BeH

and c) F2.

Answer 1

Nuclear replusive force, F = Z₁ Z₂  e² / 4 pi E_o r_AB

Given, replusive force = 15 ev

a.)

F = 15 eV

   For, H2

  F = e²  / 4 pi E_o r_AB

15 = (1.6x10^-19 ) (9x10⁹ )/ r_AB

r_{AB =} ((1.6*9)/1.5) X 10^-10   

= 9.6 X 10^-10   m   

= 9.6 Angstrom = 0.96 nm

Interatomic distance = 9.6 A^o

b.) For BeH

F = (4*1) e²  / 4 pi E_o r_AB

15 = 4*(1.6x10^-19 ) (9x10⁹ )/ r_AB

r_{AB =}   ((4*1.6*9)/1.5) X 10^-10   

= 4*9.6 X 10^-10   m   

= 4*9.6 Angstrom =38.4 Angstrom

= 3.84 nm

Interatomic distance = 3.84 nm

c.)

For F2

F = (9*9) e²  / 4 pi E_o r_AB

15 = 81*(1.6x10^-19 ) (9x10⁹ )/ r_AB

r_{AB =}   ((4*1.6*9)/1.5) X 10^-10   

= 81*9.6 X 10^-10   m   

= 81*9.6 Angstrom

= 777.6 Angstrom

= 77.76 nm

Interatomic distance = 77.76 nm