Calculate the polarizability a(lpha) of atomic hydrogen in terms of R

[cfung]

Calculate the polarizability "a(lpha)" of atomic hydrogen in terms of R

Homework Statement

A simplified model of a hydrogen atom is that the electron cloud is a sphere of radius R with uniform charge density and total charge -e (The actual charge density in the ground state is nonuniform).

For a the uniform-density model, calculate the polarizability, "a" (alpha), of atomic hydrogen in terms of R. Consider the case where the magnitude E of the applied electric field is much smaller than the electric field required to ionized the atom. (Imagine that the hydrogen atom is inside a capacitor whose uniform field polarizes but does not accelerate the atom. Consider forces on the proton in the equilibrium situation, where the proton is displaced a distance s from the center of the electron cloud (s << R in the diagram).)

Homework Equations

None provided nor hinted by the textbook but these are the only relevant equations I could think of:

Electrcfield due to spherical shell = k*(Q/r^2)

p = aE

The Attempt at a Solution

For one thing, I was never able to understand how polarizability of hydrogen can in anyway be dependent on the atom's radius R. From every way I draw out the diagram and showing any causes that could be responsible for shifting the proton's position, I could only arrive at the following conclusion: that there must be an applied electricfield. And in order for the shifted position to be static in place, the proton must also feel an opposite force contributed by the dipole electricfield which resulted from the atom's polarization.

Please guide me to the mathematical relationship between R and the polarizability.

 

[ehild]

The dipole moment of two charges both of magnitude Q and of opposite signs, and s distance apart is P=Q*s. The polarizability is defined as P/E.
The opposite charges belong to the point-like proton, and the electron cloud, assumed as a homogeneously charged sphere of radius R.
Assume an electric field of strength E. The force on the proton is Ee. It is the same on the electron cloud, but with opposite direction. As a result, the proton and the centre of the electron cloud will move away with a distance s.
Now the proton is inside the electron cloud, but not in the centre. A spherical charge distribution can be considered as a point charge in the centre, with magnitude equal to the charge inside the sphere of radius s. (You can calculate it from the volume ratio of the small sphere to the total volume of the hydrogen atom).
The two charges attract each other according to Coulomb's law.
In equilibrium, the two forces, acting on the proton (the Coulomb force and the force from the field, eE) Cancel. Determine the equilibrium distance s. With that, you get the dipole moment using Q=e.

 

[tomcue]

The polarizability of an atom is a measure of its ability to be polarized by an external electric field. In the case of atomic hydrogen, the polarizability is dependent on the atom's radius R because the size and shape of the electron cloud affects the distribution of charge and therefore the atom's response to an external electric field.

To calculate the polarizability of atomic hydrogen in terms of R, we can use the equation p = aE, where p is the dipole moment induced in the atom, a is the polarizability, and E is the applied electric field. In this case, the electric field is due to the capacitor surrounding the atom, and we can assume that it is uniform and directed towards the atom.

The dipole moment induced in the atom is given by p = qs, where q is the charge of the proton and s is the displacement of the proton from the center of the electron cloud. We can calculate s using the electric field due to a spherical shell, which is given by E = k(Q/r^2), where k is the Coulomb constant, Q is the total charge of the electron cloud, and r is the distance from the center of the electron cloud to the proton.

Assuming that the atom is in equilibrium, the force on the proton due to the electric field must be balanced by the force due to the dipole moment of the atom. This can be expressed as F = qE = p(dE/dr), where dE/dr is the gradient of the electric field.

Combining these equations, we can solve for the polarizability a in terms of R:

a = qs^2/(dE/dr) = q^2s^2/(kQ/r^2) = q^2s^2r^2/(kQ)

Substituting s = R and Q = -e, we get:

a = q^2R^2/(ke)

Therefore, the polarizability of atomic hydrogen in terms of R is directly proportional to the proton's charge q and the atom's radius R, and inversely proportional to the electric field strength E.