Emission in hydrogen atom: recoil and photon properties

In summary, the experiment setup has a single hydrogen atom in the excited state, with a magnetic field orienting the atom so that precession of its total angular momentum points to the +z-axis. When the atom is in this state, it relaxes to the ground state, emitting a photon with a wavelength of 1.21567 angstroms.
  • #1
deccard
30
0
I have been trying to picture the whole process of a photon emission by a atom. So to have good understanding what is going on, I have came up with following experimental setup. A single hydrogen atom in excited state [tex]^2\!P_{1/2}[/tex], which has been orientated with a magnetic field so that precession of its total angular momentum is pointing to +z axis. The hydrogen atom is located at rest in the middle of a hollow spherical detector with radius [tex]r=1cm[/tex].

Now, without of the presence of any external magnetic field, the excited state is relaxed to the ground state in the dipole transition [tex]^2\!P_{1/2}\rightarrow^2\!\!S_{1/2}[/tex] with an emission of a photon with wavelength [tex]\lambda=1 215.674\textrm{\AA}[/tex].

Okay now to the questions:

1.
As the emitted photon has momentum [tex]p=h/\lambda[/tex], we get that the hydrogen atom is given recoil [tex]p=h/\lambda=mv[/tex]. So the time between the detection of the photon and detection of the hydrogen atom at the detector is

[tex]t=\frac{r\lambda m}{h}=3.1\textrm{ms}[/tex],

where [tex]m[/tex] is the mass of hydrogen atom.

right?

2.
Because [tex]J=j=l+s=1/2[/tex], the state [tex]^2\!P_{1/2}[/tex] can have following configurations with notation[tex]\right \left| l,m_l,m_s\rangle[/tex]
[tex]\right \left| 1,+1,-1/2\rangle[/tex]
[tex]\right \left| 1,-1,+1/2\rangle[/tex]
[tex]\right \left| 1,0,+1/2\rangle[/tex]
[tex]\right \left| 1,0,-1/2\rangle[/tex]

and so, as the spin doesn't change, the transition can be one of these

[tex]\right \left| 1,+1,-1/2\rangle \rightarrow \right \left| 0,0,-1/2\rangle[/tex]
[tex]\right \left| 1,-1,+1/2\rangle \rightarrow \right \left| 0,0,+1/2\rangle[/tex]
[tex]\right \left| 1,0,+1/2\rangle \rightarrow \right \left| 0,0,+1/2\rangle[/tex]
[tex]\right \left| 1,0,-1/2\rangle \rightarrow \right \left| 0,0,-1/2\rangle[/tex]

right?

3.
Is the reason for the unchanging spin the need of [tex]\Delta j=\pm1[/tex]? Photon has spin [tex]\pm1[/tex] and in dipole transition we must have [tex]\Delta l=\pm1[/tex]. Which would mean that we are left with [tex]\Delta s=0[/tex].

I will continue my questions when these are answered.
 
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  • #2
1. Seems fine with me. Kind of a contrived situation, though.

2. Seems fine.

3. Right. (assuming LS-coupling)
 
  • #3
Now, I am little bit confused. I have started to doubt the permissibility of the two transitions below. (With [tex]\right \left| l,m_l,m_s\rangle[/tex] )

[tex]
\right \left| 1,0,+1/2\rangle \rightarrow \right \left| 0,0,+1/2\rangle
[/tex]
[tex]
\right \left| 1,0,-1/2\rangle \rightarrow \right \left| 0,0,-1/2\rangle
[/tex]

In the transition [tex] ^2\!P_{1/2}\rightarrow^2\!\!S_{1/2} [/tex] we have [tex]\Delta J=0[/tex], so a transition which has [tex]\Delta m_{j}=0[/tex] cannot occur. Can I calculate [tex]\Delta m_{j}[/tex] as

[tex]\Delta m_{j} = m_{ji}-m_{jf}=m_{li}+m_{si}-(m_{lf}+m_{sf})[/tex]

where indices i f represent initial and final states respectively?

If I can, this would give for the transitions [tex]\Delta m_{j}=0[/tex], which would then make the transitions forbidden.

On the other hand in the transition we have [tex]\Delta l=1[/tex] and [tex]\Delta m_{l}=0[/tex], which should be enough to make the transition allowed.

So... If some one could elaborate this little more.
 

Related to Emission in hydrogen atom: recoil and photon properties

1. What is emission in a hydrogen atom?

Emission in a hydrogen atom refers to the process by which an electron transitions from a higher energy level to a lower energy level, releasing energy in the form of a photon.

2. How does recoil affect emission in a hydrogen atom?

Recoil is the backward movement of an atom due to the release of a photon. In a hydrogen atom, recoil can affect the energy and direction of the emitted photon.

3. What factors determine the properties of the emitted photon in a hydrogen atom?

The properties of the emitted photon, such as its wavelength and energy, are determined by the energy difference between the initial and final energy levels of the electron, as well as the quantum numbers associated with these levels.

4. What is the relationship between emission and absorption in a hydrogen atom?

Emission and absorption are opposite processes in a hydrogen atom. In emission, an electron releases energy in the form of a photon while transitioning to a lower energy level. In absorption, an electron absorbs energy from a photon and transitions to a higher energy level.

5. Can emission in a hydrogen atom be controlled or manipulated?

Yes, the emission of photons in a hydrogen atom can be controlled or manipulated by changing the energy levels of the electron through external factors such as temperature or electric fields.

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