Saved Bookmarks
| 1. |
The natural angular frequency of vibration of an HF molecule is 7.79xx10^(14) rad/s. There are 13 rotational levels between the zero level and the first excitation level. Estimate distance between the centre of the atoms in this molecule. |
|
Answer» `epsi_(l)=epsi_(0)^(vib)+epsi_(l)^(ROT)=(homega)/2+(l(l+1)h^(2))/(2J)`, where l = 1, 2,........, 13 It is clear from the diagram that the fourteenth vibration-rotational level coincides with the first purely vibrational level: `epsi_(14)=(homega)/2+(14xx15h^(2))/(2f)=(3homega)/2,"where "(14xx15h^(2))/(2J)=homega` This makes it possible to determine the molecule.s MOMENT of inertia about its centre of mass: `J=105h//omega`. On the other hand, the moment of inertia is `J=m_(H)r_(H)^(2)+m_(F)r_(F)^(2),"where "r_(H)+r_(F)=d` is the distance sought between the centres of atoms and `m_(H)r_(H)-m_(F)r_(F)`. But `m_(F)-19m_(H)`. Therefore `r_(H)" "19r_(F)" "19d//20`. Substituting into the expression for the moment of inertia, we obtain `J=m_(H)r_(H)(r_(H)+r_(F))=19/20m_(H)d^(2)` For the distance between the centres we obtain `d=sqrt((2.1xx10^(3)h)/(19omegam_(H)))` 
|
|