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The force constant is
Check Your Understanding The vibrational frequency of the hydrogen iodide HI diatomic molecule is $6.69\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\text{\hspace{0.05em}}13}\phantom{\rule{0.2em}{0ex}}\text{Hz}$ . (a) What is the force constant of the molecular bond between the hydrogen and the iodine atoms? (b) What is the energy of the emitted photon when this molecule makes a transition between adjacent vibrational energy levels?
a. 295 N/m; b. 0.277 eV
The quantum oscillator differs from the classic oscillator in three ways:
First, the ground state of a quantum oscillator is ${E}_{0}=\hslash \omega \text{/}2,$ not zero. In the classical view, the lowest energy is zero. The nonexistence of a zero-energy state is common for all quantum-mechanical systems because of omnipresent fluctuations that are a consequence of the Heisenberg uncertainty principle. If a quantum particle sat motionless at the bottom of the potential well, its momentum as well as its position would have to be simultaneously exact, which would violate the Heisenberg uncertainty principle. Therefore, the lowest-energy state must be characterized by uncertainties in momentum and in position, so the ground state of a quantum particle must lie above the bottom of the potential well.
Second, a particle in a quantum harmonic oscillator potential can be found with nonzero probability outside the interval $\text{\u2212}A\le x\le +A$ . In a classic formulation of the problem, the particle would not have any energy to be in this region. The probability of finding a ground-state quantum particle in the classically forbidden region is about 16%.
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