Why is the peak of the graph more for greater wavelengths? Since wavelength is inversely proportional to energy and relative intensity, shouldn't the peak for greater wavelength be lesser? Please explain the following graph of compton scattering.
[Physics] Compton Effect (graph between relative intensity and wavelength)
intensityphotonswavelength
Best Answer
The relative intensity depends more on just the wavelength, it depends on the molar extinction coefficient of the incident substance.
Beer's law simply states that the transmittance or absorbption of a substance is dependent on the initial intensity or power. The measurement of the intensity after such a transmittance factor gives you the relative intensity.
The law looks like this: $ 2 - \log(\frac{I}{I_{0}}) = T $ where $ T $ is a decimal percentage as a multiplying factor against $ I_{0} $ itself.
You can see that if we took this literally and tried to create iterative profiles for the transmittance we would arrive at some absorption coefficient that is related to some property of the substance.
This property in known as the molar extinction coefficient and is measured from absorption $ A $ by:
$ A = \epsilon c l $
This is the main factor in relative intensity profiles through substances, not the wavelength. Although the trend you are enquiring about is a variable in Compton scattering.
$ (\lambda_{2} - \lambda_{1}) = \frac{m_{e}}{h c}(1-\cos(\theta)) $
The intensity of one electron's world-line can be estimated using a linear expression for Thomas precession:
$ I_{e_{-}} = \frac{m_{e}}{h c} $
Generalizing this to the current density $ \hat{j} $ of the light beam of your apparatus, usually given in Watts with a known aperture, you can substitute the intensity of the beam into the factor for the scattering angle $ (1 - \cos(\theta)) $
As the number of electrons $ N_{e} $ can be determined from $ \hat{j} $:
$ \hat{j} = N_{e}qA\hat{v} \Rightarrow N_{e} = \frac{j}{qAc} $ for $ c \sim v $
Now, for an incident intensity $ I_{0} = N_{e}\frac{m_{e}}{h c} $ and known incident wavelength $ \lambda_{m} $ we can calculate the scattering angle or $ \Delta\lambda $ for a measurement you make of either the angle or reflected wavelength, using:
$$ (\lambda_{m} - \lambda_{0}) = N_{e}\frac{m_{e}}{h c}(1-\cos(\theta)) $$
As you can see from the graphs above, The optimal scattering angle for conversation of energy is $ \theta = 45^{\circ} $ as the measured intensity is very close to when the light was not scattered at all. By looking at the scattering angles of greater than $ 90^{\circ} $, we can see that the reflected wavelength has changed considerably.
An observation that could be made is that the magnitude of change is related to the critical points of a standardly modelled wave-equation.
The phase of a wave is $ \psi(\phi) = e^{i\phi} \Rightarrow \cos(\phi) + i \sin(\phi) $ with $ \phi $ chosen for the nature of wave, examples would be $ \frac{2\pi nx}{L} $ and $ \frac{2 \pi t}{T} $ or in general $ kz - \omega t $.
For a before and after angle of $ \phi = 45^{\circ} $, the periodic identity $ \sin(\phi + \frac{\pi}{4}) = \cos(\phi) $ is used, as there is less of a need for a change in wavelength, as there is an innate resonance for that phase difference in harmonic motion.