This course introduces a number of fundamental physical principles which are used to predict, describe and analyse the behaviour of molecular systems.

**Course Objectives**

By the completion of the course you should be able to:

- solve problems of linear and rotational molecular motion using Newton's Laws;
- solve problems involving molecular collisions using the conservation of linear and angular momentum;
- apply the principles of the conservation of energy to molecular dynamics;
- calculate the electrostatic field, force, potential and potential energy for an assembly of charges;
- sketch the field lines and equipotentials for such an assembly;
- solve problems involving the motion of charged particles in uniform magnetic fields;
- explain the conditions required for a diatomic molecule to execute simple harmonic motion;
- qualitatively explain the origin and transmission of electromagnetic waves
and its interaction with molecules;
- determine the centre of mass of a molecule.

**Key Concepts**

Forces, fields, potential energy, potential, conservation of linear and angular momentum, conservation of energy.

**Important Dates **

Chemistry test - Friday 12^{th} January 1996, 2 - 4.45 pm.

Physical Chemistry examination - provisionally Monday 17th June, am.**Course
Outline**

- Introduction to the first year Physical Chemistry Course
- Newton's Laws applied to atoms and molecules
- Molecular collisions - applying the conservation of linear momentum
- Origins of inter-molecular forces - the electrostatic force
- Dipolar molecules - calculating forces by the principle of superposition
- Electrostatic fields and their visualisation - the dipole field
- Electrostatic work and potential energy - molecular dissociation
- Electrostatic potential, potential difference and its visualisation - the dipole potential
- Kinetic energy and the conservation of energy - how can reactions occur?
- Molecular vibrations - simple harmonic motion by the conservation of energy
- Magnetic field and magnetic force - mass spectrometry
- Electromagnetic waves and their interaction with molecules
- The dynamics of rotating molecules
- Coupled oscillation and rotation - the conservation of angular momentum.

"The Principles of Physics", Frank J. Blatt, 3rd edition, Allyn and Bacon.

*Cartesian Coordinates:* *Cylindical Coordinates:*

*Linear (constant a):*

*Rotational (constant α)*

__Question__

Answer part (a) and EITHER (b) OR (c). If you answer both (b) and (c) only the first one you have answered will be marked.

(a) Shown below is a graph of the potential energy, *U*, as a function of
interatomic separation, *r*, of two, different homonuclear diatomic
molecules, *X2* and *Y2*.

(i) What is the equilibrium separation of the atoms in *X2*?

[2 marks]

(ii) For what range of separations is the force between the atoms in *X2*
repulsive?

[2 marks]

(iii) What is the minimum energy needed to dissociate the molecule *X2*?

[2 marks]

(iv) If a photon with energy 6 eV is absorbed by the molecule *X2*, what
is the final kinetic energy of each atom? You should assume that the molecule
is at rest, i.e. it is not vibrating or rotating. State any further assumptions
you have made.

[5 marks]

(v) Which molecule has the greater bond stiffness for small perturbations from equilibrium? Give a brief (1 paragraph) explanation of your reasoning.

[4 marks]

(b) A hydrogen atom travelling with initial velocity
*u*î strikes a hydrogen molecule which is at rest, as
shown below. After it has struck, the molecule has a velocity
*v*î and the atom is at rest

Assuming that there is no interaction potential between the atom and molecule, show that the molecule must be oscillating with an amplitude,

where κ is the bond stiffness of the hydrogen molecule.

[7 marks]

(c) An ion of charge *q* and mass *m* travels in a region of uniform
magnetic field __B__. With __B__ perpendicular to __v__, where
__v__ is the velocity vector of the ion, explain why the ion will follow a
circular path and derive an expression, involving the variables referred to
above, for the radius of the circular path.

[7 marks]

__Answers__

(a) (i) Equilibrium separation of the atoms *X* occur where the potential
energy is at a minimum. Measuring from the graph this gives an equilibrium
separation of 1Å.

(ii) The force is given by the negative of the gradient of the potential
energy, i.e. *dU = -*F*.d*r, hence the
interatomic force is repulsive wherever the gradient of U is positive. From the
graph this is true for r < 1Å, and r cannot be less than 0Å.

(iii) The minimum dissociation energy can be found if we assume that at infinite separation the H atoms have zero kinetic energy (this can only occur if they separated by moving in exactly opposite directions). Then the dissociation energy is simply the difference in potential energy at equilibrium and as r tends to infinity. From the graph this means the dissociation energy is 4eV.

(iv) If we assume that the photon is absorbed and causes the atoms to fly
apart in opposite directions, as in (iii), then the excess kinetic energy is
2eV. Since the molecule is homonuclear each *X* has the same mass and by
the conservation of momentum each must carry equal but opposite momentum, i.e.
they have the same kinetic energy. The kinetic energy is therefore 1eV for each
atom.

In fact the photon has a momentum given by the de Broglie relationship, p = h/λ, which must be conserved after the absorption. This gives the molecule, initially at rest, a very small component of velocity in the direction of the photon's flight path and this may result in a small difference in the kinetic energy of each atom. For example if the atoms dissociate along the same line as the photon followed the one travelling in the same direction as the photon will have slightly more kinetic energy than the one travelling in the opposite direction.

(v) The bond stiffness of *X2* is greater than that of *Y2*. This
can be seen if we consider a small extension / compression about equilibrium.
For the same displacement the potential energy in *X2* has risen by a
greater amount than *Y2 *and hence the energetic cost for a small
displacement is greater, i.e. the bond is stiffer.

(b) The hydrogen atom travelling at velocity
*u***î** has a kinetic energy

where m is the mass of a hydrogen atom. Since there is no interaction
potential between atom and molecule and the molecule is at rest, the initial
energy of the system is simply this kinetic energy. Subsequently the atom is at
rest and the molecule has a linear velocity *v***î**
giving it a kinetic energy of

The energy taken up in the molecule's oscillations is given by the difference in the initial and final kinetic energy of the system. We can find this by determining v in terms of u by conservation of linear momentum

and then total energy of the oscillations, *Eosc* is given by

To find the amplitude, A, of the oscillations we simply note that during simple harmonic motion the kinetic energy is zero when the molecule is at its maximum and minimum extension and that the potential energy is at this point given by

which after re-arrangement gives

and hence

as required.

(c) The force, F, acting on the ion is the Lorentz force given by

This means that the force always acts at right angles to the plane defined by
the instantaneous velocity and magnetic field vectors. Since the force is
always perpendicular to the direction of motion, the magnitude of the velocity
cannot be altered (no work is being done since
F*.d*s* *= 0 in this case). By definition
this means that the motion must be circular and the acceleration on the
particle is centripetal.

Hence we can equate the magnitude of the Lorentz force to the product of the
centripetal acceleration and the particle's mass, *m*.

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