- #1

deccard

- 30

- 0

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.