- #1
deccard
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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.
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.