Question
What value of the repulsion constant, $n$, gives the measured dissociation energy of 221 kcal/mole for NaF?
Step 1
We know that 1 kcal/mole is approximately equal to 0.0434 eV, so we multiply 221 kcal/mole by 0.0434 eV to get approximately 9.583 eV. \[E = 221 \, \text{kcal/mole} \times 0.0434 \, \text{eV} = 9.583 \, \text{eV}\] Show more…
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