Knowledge Bank

In the context of the Born-Oppenheimer approximation, one should expect the electric field gradient at a nuclear site to depend on the inter-nuclear distance. Expanding this function about the equilibrium distance and averaging over rotation-vibration would then give. $$q(v,J)=q_{o} +q_{1}{<\zeta>} + q_{2} {\zeta^{2}} +.....$$ where $\zeta =(r-re)/re$. We have evaluated the averages after the manner of Dunham’s energy expansion, and find $$ <\zeta^{n}>=\sum_{Im}{Z}^{n}{Im}(v+1/2)^{1}[J(J+1)]^{m}. $$ where, to second order in [FIGURE] $$ \begin{array}{ll} Z^{1}{00}= (-15a_{3}/4 + 23a_{1} a_{2}/4 - 21a_{1}^{3}/8)(B_{e}/\omega_{e})^{2} &Z^{1}{1}0 = -3a{1} (B_{e}/\omega_{e})\ Z^{1}{20}= (-15a{3} + 39a_{1}a_{2} - 45a_{1}^{3}/2)(B_{e}/\omega_{e})^{2} & Z^{1}{01} = 4(B{e}/\omega_{e})^{2}\ Z^{2}{00}=(-3a{2}/2 + 7a_{1}^{2}/4)(B_{e}/\omega_{e})^{2} & Z^{2}{10} = 2(B{e}/\omega)\ Z^{2}{20}=(-6a{2} + 15a_{1}^{2})(B_{e}/\omega_{e})^{2} & Z_{00}^{3}=(-7a_{1}/4)(B_{e}/\omega_{e})^{2}\ Z^{3}{20}= - 15a{1}(B_{e}/\omega_{e})^{2} & Z_{00}^{4} = (3/2)(B_{e}/\omega_{e})^{2}\ Z_{20}^{4} = 6(B_{e}/\omega_{e})^{2}. \end{array} $$ This prediction is tested against experimental data for HCl and KCl. This work is supported in part by a Northwest Area Foundation Grant of Research Corporation.