Question (1740001)

Energy and amplitude of vibration of HCl

HCl is a diatomic molecule that can be modeled as two balls connected by a spring, i.e. a quantum oscillator. Experiments indicate that the spring constant k of the bond is 480 N/m. Assume the reduced mass µ of the molecule to be about 1.6 × 10−27kg. Treat the molecule as a quantum oscillator of stiness k and mass 1.6 × 10−27kg.

(a) Determine the energies of the two lowest-energy vibrational states, in units of eV. (That's the ground state and rst excited state.) (b) For each energy state, determine the amplitude of vibration of the oscillator if you treat it classically with an energy E = 1/2kA2. (c) What energy photon, in units of eV, is emitted if the molecule transitions from the rst excited vibrational state to the ground state? (d) What is the change in the amplitude of vibration if the molecule transitions from the rst excited vibrational state to the ground state? (e) After the molecule emits the photon, it loses energy. Does its frequency of oscillation increase, decrease, or remain constant? Explain your answer.

Solution

(a) Treat the HCl molecule as two particles attached by a spring.

Figure 1: Ball-spring model of a diatomic molecule.

Both particles are oscillating. Therefore, it is not a quantum oscillator which consists of a single particle attached to a spring. However, we can model the diatomic molecule as a quantum oscillator if we use the reduced mass µ of the molecule, as shown below. Note that a chlorine atom is much more massive than a hydrogen atom. As a result, the chlorine atom nearly acts as a rigid barrier (since its kinetic energy will be negligible and it will have a small amplitude oscillation). As a result, the reduced mass of the system is actually very close to the mass of the hydrogen atom itself. The vibrational energy of a quantum oscillator is

E = N¯hω0 + E0 where N = 0, 1, 2, . . . Figure 2: A model of a diatomic molecule as a quantum oscillator.

and

E0 = 1/2¯hω0

p The angular frequency of the oscillation depends on the stiness and mass of the oscillator, ω0 = k/m. We must use the reduced mass of the system though it is close to the mass of the hydrogen atom itself. The lowest two energy states are for N=1 and N=2.

p E0 = 1/2¯h k/m s 480 N/m = 1/2(6.58 × 10−16 eV s) 1.6 × 10−27 kg = 1/2(6.58 × 10−16 eV s)(5.48 × 1014 rad/s) = 0.18 eV

E1 = 3/2¯hω0 = 0.54 eV

(b) The energy of a classical harmonic oscillator is related to its amplitude. Using this expression, we can calculate the amplitude of oscillation of the reduced mass in the model. (Note: that in the real molecule, both the hydrogen atom and the chlorine atom oscillate.) Be sure to convert the energy to joules so that the amplitude will be in meters.

E = 1/2kA2 r 2E A = k r 2(0.18 eV)(1.6 × 10−19 J) A = 0 480 −11 A0 = 1.1 × 10 m r 2(0.54 eV)(1.6 × 10−19 J) A = 1 480 = 1.9 × 10−11 m

Note that the higher energy state has a greater amplitude of vibration. Also, the amplitude in each case is about 10 times smaller than the diameter of the hydrogen atom itself which is roughly 1 × 10−10 m which makes sense because we expect small oscillations for the atoms. Note that this is not the amplitude of vibration of the hydrogen atom or the chlorine atom, but rather it's the amplitude of the single particle in our model that has a mass equal to the reduced mass of the system.

(c) If the oscillator transitions from N=1 to N=0, it emits a photon with an energy given by the energy principle. Consider the oscillator and the photon as the system. Neglect the rest energy of the oscillator since it does not change during the process because the mass of the molecule remains constant. There is no work done on the system, and no heat is transferred to the system during the process.

Ef = Ei + W + Q

Ef = Ei + 0 + 0

Ef = Ei

Dene the initial state to be the oscillator at N=1. Dene the nal state of the system to be the oscillator at N=0 and the photon with kinetic energy.

Ef = Ei

E0 + Ephoton = E1

Solve for the energy of the emitted photon.

Ephoton = E1 − E0 = 0.54 eV − 0.18 eV = 0.36 eV

(d) The change in amplitude is 1.9×10−11 m−1.1×10−11 m = 8×10−12 m. This means that the amplitude decreased by

∆A 8 × 10−12 m = −11 A1 1.9 × 10 m = 42%

due to the emission of the photon. p (e) The frequency of oscillation is ω0 = k/m which depends on the stiness and mass of the oscillator. Though the energy of the oscillator is E = N¯hω0 + E0, the frequency does NOT change for higher energies. It remains constant. Only N changes for higher energies. Therefore, though the oscillator emits a photon and loses energy, its frequency stays the same.

This is an important point worth understanding, the frequency of an oscillator depends on the physical qualities of the oscillatorstiness and massand is independent of the energy of the oscillator. As an oscillator's energy changes, its amplitude of vibration changes.