(a) calculate the electric potential energy for a K+ and Br- ions separated by 0
ID: 2109685 • Letter: #
Question
(a) calculate the electric potential energy for a K+ and Br- ions separated by 0.29nm (the equilibrium size of the KBr molecule). Treat the ions as point charges. (b) The ionization energy for K atom is 4.3 eV. Br has the electron affinity of 3.5 eV (this is how much lower the energy of an extra electron is in the Br 4p subshell compared to the free electron). Use these data and the results of part (a) to estimate the binding energy of the KBr molecule, in eV/molecule and in J/mole. Do you think the actual binding energy is higher, lower or equal to this estimate (explain why)?
Explanation / Answer
CHARGE ON k+ =1.602*10^-19
CHARGE ON BR-=-1.602*10^-19
ELECTRIC POTENTIAL=KQ1Q2/R
WHERE R=0.29*10^-9
V=9*10^9 * 1.602*10^-19 * 1.602*10^-19/0.29*10^-9
V=2.56*10p^-22 V
IONISATION ENERGY OF K=4.3
ELECTRON AFFINITY OF BR=3.5
TOTAL=4.3+3.5=7.8 EV
CONVERT POTENTIAL TO eV AND SUBSTRACT WITH rhs TO GET BINDING ENERGY
SOME MASS LOSS OCCURS TO CONVERT MASS TO ENEGRY...
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