Get into the wall

1. Jul 8, 2006

Neelesh

Get into the wall !!!

I want to say that we can't go into the wall b'cus we know that we are made up of atoms (mainly electrons, protons and neutrons) and all these are fermions, and so as wall. Since fermions follow Pauli's Exclusion Principle. Suppose, we get inside the wall so the fermions (electrons, protons and neutrons) occupy the same quantum state and that is the violation Pauli's Exclusion Principle. So, this implies that we can't enter inside the wall.

Now, photons can enter inside the glass slab but not inside a wall (both are solids)?
Answer is that since photons are bosons(don't obey Pauli's Exclusion Principle) so, when they enter inside the glass the interact with fermions which is not the violation of Pauli's Exclusion Principle. But, the photons of visible light don't have sufficient energy according to equation-

E = hc/(wavelength)

So, it can't pass through wall but instead a gamma ray is taken (havin' wavelength > wavelength of visible light, can enter inside a wall.

So, is this proposition true?

2. Jul 8, 2006

Gokul43201

Staff Emeritus
No. Here's a counter-example: water is highly transparent at visible energies (~10^15 Hz) but highly opaque at about 10 times higher energies (~10^16 Hz) and 10 times lower energies (~10^14 Hz).

Ref: See fig 7.9, p 315, J. D. Jackson, Classical Electrodynamics (3 Ed.)

3. Jul 8, 2006

Creator

Radio waves have energy far less than that of visible light and yet they do get through the wall.:uhh:

The reason is:
Coefficients of absorption vary both with type of material and specific wavelength, the value of which must usually be determined by experiment.

However...
For any particular material and wavelength, the absorbance varies with concentration of the absorbing material, the length of material the light passes through, and the coefficient of absorption.

Here, have a Beer:
http://en.wikipedia.org/wiki/Beer-Lambert_law

Creator

Last edited: Jul 8, 2006