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Emission in hydrogen atom: recoil and photon properties

  1. Dec 10, 2009 #1
    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.
     
  2. jcsd
  3. Dec 10, 2009 #2

    alxm

    User Avatar
    Science Advisor

    1. Seems fine with me. Kind of a contrived situation, though.

    2. Seems fine.

    3. Right. (assuming LS-coupling)
     
  4. Dec 12, 2009 #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.
     
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