(n=3, l=2, j=5/2) -> (n=2, l=1, j=1/2) hydrogen transition?

In summary, the book Physics of Atoms and Molecules by Bransden & Joachain explains that the selection rule for the quantum number l in the dipole approximation remains \Delta l = \pm 1, and the selection rule for the quantum number j is \Delta j = 0, \pm 1. The transition with \Delta j = \pm 2 is not mentioned, and the relationship between j and l is j = l \pm \frac{1}{2}.
  • #1
andresordonez
68
0
Hi, I have a doubt about the fine structure of the hydrogenic atoms. In the section 3.2 of the book Physics of Atoms and Molecules by Bransden & Joachain says:

Since the electric dipole operator [tex] \mathbf{D} = -e\mathbf{r} [/tex] does not depend on the spin, the selection rule derived in Chapter 4 for the quantum number [tex] l [/tex] (in the dipole approximation) remains

[tex] \Delta l = \pm 1 [/tex]

from which it follows that the selection rule with respect to the quantum number [tex] j [/tex] is

[tex] \Delta j = 0, \pm 1 [/tex]

My question is, why [tex] \Delta j = \pm 2 [/tex] is not mentioned here? For example, why the transition
[tex] (n=3, l=2, j=5/2) \rightarrow (n=2, l=1, j=1/2) [/tex] is not considered?

(where [tex]n[/tex] is the energy number, [tex]l[/tex] is the orbital number, and [tex]j[/tex] is the total angular momentum number)
 
Physics news on Phys.org
  • #2
Why would it be ? How are j and l related ?
 
  • #3
Yeah, sorry I should have posted it before; [tex] l [/tex] and [tex]j[/tex] are related like this:

[tex] j = l \pm \frac{1}{2} [/tex]
 

FAQ: (n=3, l=2, j=5/2) -> (n=2, l=1, j=1/2) hydrogen transition?

1. What is the significance of the (n=3, l=2, j=5/2) -> (n=2, l=1, j=1/2) hydrogen transition?

The (n=3, l=2, j=5/2) -> (n=2, l=1, j=1/2) hydrogen transition refers to the energy level transition of a hydrogen atom from the excited state with quantum numbers n=3, l=2, j=5/2 to the ground state with quantum numbers n=2, l=1, j=1/2. This transition is significant because it emits a photon with a specific wavelength, which can be detected and studied to understand the properties of hydrogen atoms.

2. What is the energy difference between the (n=3, l=2, j=5/2) and (n=2, l=1, j=1/2) states in hydrogen?

The energy difference between the (n=3, l=2, j=5/2) and (n=2, l=1, j=1/2) states in hydrogen is known as the energy of the transition. In this case, it is equal to 10.2 eV (electron volts). This value can be calculated using the Rydberg formula, which relates the energy of an electron's transition between two energy levels in a hydrogen atom to the Rydberg constant and the quantum numbers of the energy levels.

3. What is the origin of the (n=3, l=2, j=5/2) -> (n=2, l=1, j=1/2) hydrogen transition?

The (n=3, l=2, j=5/2) -> (n=2, l=1, j=1/2) hydrogen transition is caused by the change in electron configuration within the atom. When an electron in the n=3 energy level drops to the n=2 energy level, it releases energy in the form of a photon. This energy is equal to the difference in energy between the two levels and corresponds to a specific wavelength of light.

4. How is the (n=3, l=2, j=5/2) -> (n=2, l=1, j=1/2) hydrogen transition observed?

The (n=3, l=2, j=5/2) -> (n=2, l=1, j=1/2) hydrogen transition can be observed using various spectroscopic techniques. One method is to pass a beam of hydrogen atoms through a prism or diffraction grating, which separates the different wavelengths of light emitted during the transition. Another method is to use a spectrometer to measure the intensity of light at different wavelengths. The resulting spectrum will show a sharp peak at the wavelength corresponding to the transition energy.

5. What is the significance of studying the (n=3, l=2, j=5/2) -> (n=2, l=1, j=1/2) hydrogen transition?

The (n=3, l=2, j=5/2) -> (n=2, l=1, j=1/2) hydrogen transition is significant because it provides valuable information about the energy levels and electron configuration of hydrogen atoms. It also has applications in fields such as astrophysics and quantum mechanics, where the properties of atoms and their transitions are studied in detail. Additionally, studying this transition can help us understand the behavior of other atoms and molecules as well.

Similar threads

Replies
21
Views
656
Replies
5
Views
2K
Replies
14
Views
2K
Replies
3
Views
3K
Replies
2
Views
1K
Replies
1
Views
2K
Back
Top