Question

When an atom is placed in an external field, the nucleus moves just a tiny bit, so that it is no longer centered, until it is once again in equilibrium As a result, the atom acquires a dipole moment. For many atoms, the induced dipole moment is proportional to the applied field; these are called linear materials, or dielectrics. This dependence, called the polarizability, can be measured fairly easily in the lab. Remarkably, Gauss’s law is all we need to get a theoretical value for the polarizability of atoms that is a surprisingly good match to the experimentally measured value. Lets see how that works:

We will model the atom as a uniform sphere, radius a, of negative charge −q, with a positive point charge q in the middle. We will assume that the external field does not distort the sphere, nor pull the nucleus all the way out of the negatively charged sphere (these are certainly good assumptions for any reasonable field you might be using!) In this problem, we are treating the positive nucleus as a test charge, and the sphere of uniform negative charge as the source charge.

(a) Draw a diagram of the model with the nucleus shifted a distance d from the center, due to the applied field. Add a free body diagram to your drawing, showing the force by the applied field and the force by the field due to the negative charges. Note that we are interested in the situation where the nucleus is in equilibrium.

(b) You don’t need to re-derive the results if we already did them in class or if they are done in the text for you - if so, you may simply quote them: What is the field due to the sphere of negative charge at a distance d from the center? Use this and your free body diagram (remember, there are two equal-magnitude forces) to find the dipole moment p = qd as a function of a and the applied field E.

(c) You should have found that the dipole moment is proportional to E. The proportionality constant α = p/E is called the atomic polarizability. Based on these results, are larger atoms easier or harder to polarize than smaller atoms, or harder to polarize?

(d) The measured polarizability times k, in units of 10−30 m3 , of helium is 0.205 and of carbon is 1.67 (from the Handbook of Chemistry and Physics.) Use this data and the proportionality factor you got using Gauss’s law to estimate the radius of each of these atoms. Your calculation (if you did it right) is actually pretty close to the actual value!

Answer 1

Filed at a distance d from the center of the sphere of electron cloud

  $= \frac{qd}{4\pi \epsilon _oa^3}$ , a is the radius of the sphere ( atom)

The nucleus and the electron cloud come to equilibrium when the force due to the external field and the internal fields on the nucleus and the electron cloud are equal.

i.e.  $E = \frac{qd}{4\pi \epsilon _oa^3}$ E is the external field applied.

dipole moment of the atom

$p= qd =E 4\pi \epsilon _oa^3$

c) for larger atoms z is high and the total charge q is high and so p (dipole moment is high is high

hence $\alpha$ is high , it is easier to polarise larger toms.

d) helium k $\alpha$ = a³ = 0.205 E-30

a = 5.896E-9 m

carbon k $\alpha$ = a³ = 1.67 E-30

a = 11.864E-9m