How Is Atomic Polarizability Calculated for a Hydrogen Atom?

[Reshma]

The Charge density of an electron cloud for a Hydrogen atom is given by:
$$\rho (r) = \left(\frac{q}{\pi a^3}\right)e^{\frac{-2r}{a}}$$
Find its polarizability($\alpha$).

My work:

Dipole moment p is:
$$\vec p = \alpha \vec E$$

I need to calculate the electric field first.
The electric field is given by Gauss's law:
$$\vec E = \left(\frac{1}{4\pi \epsilon_0}\right)\frac{Q_{total}}{r^2}\hat r$$

$$Q_{total} = \int_{0}^{r} \rho (r)d\tau$$

$$Q_{total} = \int_{0}^{r} \left(\frac{q}{\pi a^3}\right) e^{\frac{-2r}{a}} 4\pi r^2 dr$$

$$Q_{total} = \frac{4q}{a^3} \int^{r}_{0} e^{\frac{-2r}{a}} r^2 dr$$

How is this integral evaluated?

 

[Galileo]

Integrate by parts. (*groan*)

 

[Reshma]

Integrate by parts. (*groan*)

Thought so, but is there an appropriate substitution for the e^() term?

 

[marlon]

Thought so, but is there an appropriate substitution for the e^() term?

This is a very easy one. You will need two steps in the integration by parts. Just start like this

$$\frac{4q}{a^3} \int^{r}_{0} e^{\frac{-2r}{a}} r^2 dr = \frac{-a}{2} \frac{4q}{a^3} \int^{r}_{0} r^2 de^{\frac{-2r}{a}}$$

marlon

 

[inha]

You don't really need one but I suppose you could go with r'=2r/a to simplify the algebra a little bit.

 

[Reshma]

Thanks, marlon and inha, I'll try it.

 

[Reshma]

OK, the integration part was pretty lengthy and I found the magnitude of the electric field of the electron cloud.
$$E_e = \left(\frac{1}{4\pi \epsilon_0}\right)\frac{q}{r^2}\left(1 - e^{\frac{-2r}{a}} \left(1 + \frac{2r}{a} + \frac{2r^2}{a^2}\right)\right)$$

This is the field of the electron cloud. The proton will be shifted from r = 0 to a point 'd' where the applied field E equals field of the electron cloud.
So,

$$E = \left(\frac{1}{4\pi \epsilon_0}\right)\frac{q}{d^2}\left(1 - e^{\frac{-2d}{a}} \left(1 + \frac{2d}{a} + \frac{2d^2}{a^2}\right)\right)$$

How do I find the dipole moment term from this equation and hence the atomic polarizability?