Photoemission in diatomic molecules

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In summary, photoemission in diatomic molecules is the process of using light energy to excite electrons in a diatomic molecule, resulting in their emission. This can occur through either photoionization or photodetachment. Photoemission occurs when a photon of light interacts with a molecule, causing an electron to be excited or removed. It is significant in understanding electronic structure, chemical bonding, and dynamics of molecules, and is commonly studied using techniques such as photoelectron spectroscopy. Additionally, photoemission contributes to our understanding of chemical reactions by providing information about energy transfer and rearrangement of atoms during a reaction.
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Diracn
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Hello !

I have a question that breaks the head to me, jejeje

In the process of photoemission, the total cross section is defined by:

[tex]
\sigma(\omega) = \frac{4\pi}{3}\alpha a_0^2 \omega \sum_{lmm_\gamma} \left| D_{lmm_\gamma}(\omega)\right|^2
[/tex]

where [tex]\alpha[/tex] is the fine-structure constant (dimentionless), [tex]a_0[/tex] is the Bohr's radius ( units meters), [tex]\omega[/tex] is the Energy of radiation (units Jules) and [tex]D_{lmm_\gamma}[/tex] are the transition coefficients expressed of the following way

[tex]
D_{lmm_\gamma}(\omega) = \sqrt{\frac{4\pi}{3}}\bigl< \psi_{lm}(k, r)\bigl| r Y_{1}^{m_\gamma}(\theta,\phi) \bigl | \psi_0(r) \Bigr>
[/tex]

where the wave function for the emitted electron is ( [tex]k = \sqrt{2E}[/tex] )

[tex]
\psi_{lm}(k, \mathbf{r}) = \frac{1}{kr}f_{lm}(kr)Y_{lm}(\theta,\phi)
[/tex]

and the wave function for the ionized orbital is

[tex]
\psi_0( r) = \frac{1}{r}\sum_{l m}\psi_{l m}(r)Y_{l}^{m}(\theta,\phi)
[/tex]

my simple question is:

What units must have each one of the involved terms so that the total cross section is expressed in barns?

Thanks !
 
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  • #2




Thank you for your question about the units involved in the total cross section for photoemission. The total cross section is a measure of the probability for a photon to be absorbed and an electron to be ejected from an atom or molecule. It is typically expressed in units of area, specifically in barns (1 barn = 10^-28 m^2). In order for the total cross section to be expressed in barns, the units of the individual terms in the equation must be adjusted accordingly.

Starting with the fine-structure constant, \alpha, it is a dimensionless quantity and does not have any units. Therefore, it does not need to be adjusted.

Next, the Bohr's radius, a_0, is typically expressed in meters. In order to convert it to barns, it must be squared and multiplied by a conversion factor of 10^-28 m^2/barn.

Moving on to the energy of radiation, \omega, it is typically expressed in joules (J). In order to convert it to barns, it must be multiplied by a conversion factor of 10^-18 J/barn.

Finally, the transition coefficients, D_{lmm_\gamma}(\omega), are dimensionless quantities and do not need to be adjusted.

In summary, the units for each term in the equation must be adjusted as follows:

\sigma(\omega) = \frac{4\pi}{3}\alpha a_0^2 \omega \sum_{lmm_\gamma} \left| D_{lmm_\gamma}(\omega)\right|^2

Units in barns:

\sigma(\omega) = \frac{4\pi}{3}\alpha (a_0^2)(10^{-28}\text{ m}^2/\text{barn}) (\omega)(10^{-18}\text{ J}/\text{barn}) \sum_{lmm_\gamma} \left| D_{lmm_\gamma}(\omega)\right|^2

I hope this answers your question. Please let me know if you need any further clarification. Best of luck with your research!
 
  • #3


Hello! Thank you for your question. In order for the total cross section to be expressed in barns, each term in the equation must be in the appropriate units. The fine-structure constant, \alpha, is dimensionless and does not have any units. The Bohr's radius, a_0, is measured in meters. The energy of radiation, \omega, is measured in joules. The transition coefficients, D_{lmm_\gamma}, are expressed in units of meters^-1/2. Therefore, in order for the total cross section to be expressed in barns, all the units must cancel out except for the units of the Bohr's radius, which is in meters. This means that the total cross section will be expressed in units of square meters, which is the unit for barns. I hope this helps clarify your question. Thank you for your interest in photoemission in diatomic molecules!
 

1. What is photoemission in diatomic molecules?

Photoemission in diatomic molecules is a process in which light energy is used to excite electrons in a diatomic molecule, causing them to be emitted from the molecule. This can occur through either photoionization, where the electron is completely removed from the molecule, or photodetachment, where the electron is detached from the molecule but remains in its vicinity.

2. How does photoemission in diatomic molecules occur?

Photoemission in diatomic molecules occurs when a photon of light with sufficient energy interacts with a molecule, causing an electron to be excited to a higher energy level or to be completely removed from the molecule. This process can be described by the photoelectric effect, where the photon transfers its energy to the electron, allowing it to overcome the binding energy of the molecule.

3. What is the significance of studying photoemission in diatomic molecules?

Studying photoemission in diatomic molecules can provide valuable information about the electronic structure, chemical bonding, and dynamics of these molecules. It can also help in understanding the behavior and properties of other molecules and materials that have similar electronic structures.

4. What techniques are commonly used for studying photoemission in diatomic molecules?

The most common technique for studying photoemission in diatomic molecules is photoelectron spectroscopy, which involves shining a beam of light onto a sample and measuring the kinetic energy and intensity of the emitted electrons. Other techniques such as mass spectrometry, infrared spectroscopy, and theoretical calculations are also commonly used to investigate the properties of photoemission in diatomic molecules.

5. How does photoemission in diatomic molecules contribute to our understanding of chemical reactions?

Photoemission in diatomic molecules plays a crucial role in understanding the mechanisms and dynamics of chemical reactions. By studying the electronic and vibrational states of molecules before and after photoemission, scientists can gain insights into the energy transfer and rearrangement of atoms during a reaction. This information can help in designing and optimizing chemical reactions for various applications.

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