Absorption

Absorption

  • Non-metallic materials may be opaque or transparent to visible light. If transparent they often appeared colored .The light absorption in this group of material can be explained by two basic mechanism

i. Electron polarization

ii. Electron energy transition

  • In electron polarization light of frequency near about the relaxation frequency is absorbed.
  • In absorption of photon there is transition of electron from valence band to empty conduction band.
  • Every non-metallic material becomes opaque at some wavelength, which depends upon the energy of band gap. For example: if band gap in the case of diamond is 5.6 ev then the diamond is opaque to radiation having wavelength \(\lambda\) calculated as,

ImoDiamond using band gap 5.6ev is opaque to radiation having wavelength 0.22\(\mu\). Photon of energy greater than band gap absorbed and this region material becomes opaque.

If \(E_g<1.8ev\), visible light is absorbed. If 1.8<\(E_g\)<3.1ev then the material is colored. If \(E_g\)>3.7ev the material is transparent to visible light.

When light beam impinged on a material surface portion of the incident beam is reflected,

$$I_T’=I_\circ’ e^-\beta x\dotsm(1)$$

\(\beta\)=linear absorption coefficient in \(mm^-1\)

X=thickness of material

This is known as Baugure’s law.

There are two type of absorption mechanism

1.Rayleigh scattering

2.Compton scattering

In Rayleigh scattering photon interact with electron’s orbit an atom and is deflected without change in frequency and photon energy. This is significant for high atomic number atom.

In Compton scattering photon losses some of its energy to electron and emitted as lower energy photon. The other mechanisms are also important but less visible for non-metallic materials. This are Tyndale effect and photoelectric effect.

fig:
fig:absorption

Above figure represents photon absorption from valence band to conduction band that has as impurity level in between band gap and emission of two photons due to multiple transition from conduction band to band gap (impurity level) and then to valence band.

Transmission:

The phenomena or transmission is similar to that of absorption and reflection. The intensity of transmitted beam through a material of thickness ‘l’ is given by,

$$I_T=I_\circ(1-R)^2 e^-\beta l\dotsm(2)$$$

\(I_\circ\)=intensity of incident beam

R=reflection\reflectivity

\(\beta\)=absorption coefficient

l=thickness (specimen’s)

fig: transmission
fig: transmission

AB represents incident light of \(I_\circ\). BC represents reflected from upper surface i.e. R\(I_\circ\). The intensity of light just entered below upper surface. BD represents intensity of refracted light at point D i.e. distance x from B, intensity is, \((I_\circ-I_\circ R e^{-\alpha x}\)) where \(\alpha\)=absorption coefficient. So decreases in intensity or \((I_\circ-RI_\circ)-(I_\circ-RI_\circ)e^{-\alpha x}\)

Reflected at point D=\(R(I_\circ)e^{\alpha x}\)

\(\therefore I_T=(I_\circ-RI_\circ)e^{-\alpha x}-R(I_\circ-RI_\circ)e^{-\alpha x}\)

$$I_T=I_\circ e^{-\alpha x}(1-R-R+R^2)$$

$$ \therefore I_T=I_\circ(1-R)^2 e^{-\alpha x}$$

  1. Transparent material appeared colored due to the specific wavelength of light selectively absorbed. The colored is a result of combination of wavelength that is transmitted.
  2. If absorption is uniform for all visible wavelengths the material appears colorless. Example: high purity inorganic glass, single crystal, diamond, sapphire etc.

References:

Callister, W.D and D.G Rethwisch. Material Science and Engineering. 2nd. New Delhi: Wiley India, 2014.

Lindsay, S.M. Introduction of Nanoscience . New York : Oxford University Press, 2010.

Patton, W.J. Materials in industry . New Delhi : Prentice hall of India, 1975.

Poole, C.P. and F.J. Owens. Introduction To Nanotechnology. New Delhi: Wiley India , 2006.

Raghavan, V. Material Science and Engineering. 4th . New Delhi: Pretence-Hall of India, 2003.

Tiley, R.J.D. Understanding solids: The science of Materials. Engalnd : John wiley & Sons , 2004.

1. Important relations:

$$h\nu>E_g$$

$$h\frac{c}{\lambda}>E_g$$

$$\lambda_{min}<\frac{hc}{E_g}$$

$$\lambda_{min}<0.22 \mu m$$

$$I_T’=I_\circ’ e^-\beta x$$

$$I_T=I_\circ(1-R)^2 e^-\beta l$$

$$ \therefore I_T=I_\circ(1-R)^2 e^{-\alpha x}$$

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