The ground-state vibrational frequency for nitrogen is 2359 cm^−1. The spacing between the lines in the rightmost set of lines corresponds to 2150 cm^−1, and the line for the transition from v=0 to v=0 is at 15.58 eV. Find the ionization energy from the minimum of the ground-state potential curve

Question

The ground-state vibrational frequency for nitrogen is [latex] 2359 cm^{−1}.[/latex] The spacing between the lines in the rightmost set of lines corresponds to [latex]2150 cm^{−1} [/latex], and the line for the transition from v=0 to v=0 is at 15.58 eV. Find the ionization energy from the minimum of the ground-state potential curve to the minimum in the ion curve.

Accepted Answer

The ionization energy from the v=0 vibrational state to the v=0 vibrational state differs from the desired quantity by the difference of the zero-point vibrational energies, given by

[latex] ΔE_{zero−point}=\frac{hν_{molecule}}{2}− \frac{hν_{ion}}{2}=\frac{hc}{2}(\widetilde{v}_{molecule} −\widetilde{v}_{ion}) [/latex]

[latex]=\frac{(6.6261 × 10^{−34} J s)(2.9979 × 10^{10} cm s^{−1})}{2}(209 cm^{−1}) [/latex]

[latex]=2.08 × 10^{−21} J=0.0130 eV [/latex]

[latex] ΔE_{e−e}= ΔE_{0−0} + 0.013 eV=15.58 eV + 0.013 eV=15.59 eV [/latex]

Question 23.15: The ground-state vibrational frequency for nitrogen is 2359 ...

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