1. Atomic Physics Quantum Physics 2002 Recommended Reading: Harris Chapter 7

2. Energy Levels in Hydrogen Every possible state of the hydrogen atom has a distinct wavefunction that is specified completely by four quantum numbers (n, l, m L, m S ). In many cases the energy levels associated with the quantum numbers m L and m S are degenerate and we can describe the states by the n and l quantum numbers alone, e.g 1s, 2p, 3p, 3d,... We know that an atom can emit characteristic electromagnetic radiation when it makes transitions to states of lower energy. An atom in the ground state cannot emit radiation but it can absorb electromagnetic radiation and make a transition to higher (excited) states. We can use the hydrogen wave functions to calculate transition probabilities for an electron to change from one state to another when it absorbs or emits radiation. These calculations show that transitions in which the angular momentum quantum number l, changes by  1 are allowed to take place, (  l =  1) these are called allowed transitions while transitions for which  l   1 are not allowed to occur and are called forbidden transitions.

3. Selection Rules The rules on how the quantum numbers can change when a transition takes place are called selection rules. We can summarise them as follows: 1) There is no restriction on the change  n, in the principle quantum number 2) the orbital angular momentum quantum number l can only change by  1. 3) The magnetic quantum number m L, can only change by 0 or  1 4) The spin quantum number m S, can change from -1/2 to +1/2 or vice versa but it does not have to change for a transition to occur  l =  1  s  p (  l = +1) is allowed but s  d (  l = +2) is forbidden  p  s (  l = -1) and p  d (  l = +1) are allowed  d  p (  l = -1) and d  f (  l = +1) are allowed

4. Energy Level Diagram of Hydrogen

5. Atom in a Magnetic Field In the absence of a magnetic field we can (in principle) find the energy levels of an atom by solving the Schrodinger equation for the hydrogen atom or hydrogen like ion (He +, Li 2+, Be 3+ …) while for a multielectron atom the potential energy is much more complicated where the first term is the interaction between each electron and the nucleus (with charge +Ze), while the second term represents the Coulomb repulsion between each pair of electrons. r1r1 r2r2 r3r3 r 1 - r 2

6. In general, in the absence of a magnetic field we can write the Schrodinger equation in the following form where is the total energy operator (the Hamiltonian operator) and E n are the allowed energy levels for the electrons in the atom. If we now place the atom in a magnetic field then we will have an additional term in the potential energy due to the interaction between the magnetic moment of the atom and the magnetic field. This magnetic interaction energy is given by where  is the magnetic moment of the atom due to orbital and spin angular momentum and B is the applied magnetic field. The total energy operator then becomes

7. We have seen that the magnetic moment is proportional to the angular momentum, So, for an atom with total angular momentum J, the magnetic moment operator is where g j is a proportionality constant called the Lande g-factor (see below).   B  z z If the magnetic field B, is along the z-axis then the Interaction Energy is and If we operate on the wavefunction with this operator we obtain butso 

8. This shows us that when an atom is placed in a magnetic field B, the energy levels are shifted and the angular momentum degeneracy is removed, that is, the energy levels are split. For example, if J = 1 then m j = -1, 0, 1 and this energy level will be split into three when a magnetic field is applied Note that the size of the splitting depends on the strength of the magnetic field B No fieldApplied field B

9. Lande g-factor The magnetic moment is related to the angular momentum, by where g j is a proportionality constant called the Lande g-factor. This factor depends on the state the atom is in and is given by Example 1: If the atom is in a 1 D 2 state, then L = 2 (D-state), J = 2, and S = 0 (since 2S+1 = 1) and the g-factor for this state is Example 2: If the atom is in a 2 P 1/2 state, then L = 1 (P-state), J = 1/2, and S = 1/2 (since 2S+1 = 2) and the g-factor for this state is

10. Lande g-factors for one electron atoms

11. Zeeman Effect We have seen that when an atom is placed in a magnetic field the energy levels are split, i.e. Suppose that in the absence of a magnetic field, we study the transition between two energy levels of an atom In the presence of a magnetic field, both of these levels may split, i.e. and So when the atom is placed in a magnetic field the energy of the emitted photon is

12. or, in terms of the frequency or where is the Larmor frequency In a magnetic field the single emission line due to transitions between the two states will now split into a series of lines. This splitting is called Zeeman Splitting and the effect is known as the Zeeman Effect There are two cases we have to consider: CASE 1: Suppose that the atom has an even number of electrons in its outer shell so that S = 0, that is, we are looking at transitions between two singlet states. Then, since J = L + S  J = L for singlet states and the Lande g-factor is for all singlet states

13. then, the expression for the transition frequencies becomes where  m j = m j ’’ - m j ’ From the selection rules  m j = 0,  1, so we are left with the following frequencies. the single emission line is split into three lines when the atom is placed into a magnetic field. This is best illustrated by an example. The cadmium atom has an emission line at a wavelength of 643.8 nm in the absence of a magnetic field. This emission line is due to a transition between the 6 1 D 2 state and the 5 1 P 1 state of the atom. 6 1 D 2 state  n = 6, L = 2, J = 2, S = 0 and m j = -2, -1, 0, 1, 2 5 1 P 1 state  n = 5, L = 1, J = 1, S = 0 and m j = -1, 0, 1

14. Energy Level Diagram for 1 D 2  1 P 1 transition in Cd  m j = -1  m j = 0  m j = +1 mjmj 2 1 0 -2 1 0 6 1 D 2 5 1 P 1  No B field With magnetic field B Polarisation 

15. Polarization of Emitted Lines If we observe the polarisation of the emitted lines when the atom is placed in a magnetic field, then it is found that the lines are polarised. For lines which have:  m = 0, the emitted light is polarised parallel to the direction of the magnetic field. These lines are referred to as  -lines or  -polarised lines  m =  1, the emitted light is polarised perpendicular to the direction of the magnetic field. These lines are referred to as  -lines or  - polarised lines B B Polariser  -polarised  -polarised

16. No B field With magnetic field B The splitting between the lines is  = L, i.e. Note that the splitting is directly proportional to the strength of the magnetic field, if we increase B then the splitting increases. If we place the atom in a known magnetic field, then we can obtain the ratio of e/m from the splitting of the lines When a spectral line splits into three lines in the presence of a magnetic field, the effect is known as the Normal Zeeman Effect Very accurate method to determine e/m ratio. Vary B, and measure splittings. plot graph of  versus B get e/m from slope

17. We now consider the more general case where S  0, and the transition frequencies are given by Case 2: Anomalous Zeeman Effect Here the upper and lower levels can have different g-factors and the line can split into more than three components. For example consider the following transition in sodium 2 P 3/2  2 S 1/2 (this is the D 2 -line that gives a sodium lamp its characteristic yellow colour) Lande g-factors: The 2 P 3/2 line has J = 3/2, L = 1, S = 1/2 and m j = -3/2, -1/2, 1/2, 3/2 The 2 S 1/2 line has J = 1/2, L = 0, S = 1/2 and m j = -1/2, +1/2 The 2 P 3/2 state splits into four levels while the 2 S 1/2 state splits into two levels when a magnetic field is applied

18.  m j = -1  m j = 0  m j = +1 mjmj 3/2 1/2 -3/2 -1/2 1/2 -1/2 2 P 3/2 2 S 1/2 No B field With magnetic field B Polarisation  6/3 2/3 -6/3 -2/3 gjmjgjmj 1  g=2, L = 0, S = 1/2, J = 1/2 g=4/3, L = 1, S = 1/2, J = 3/2 153264 

19. Frequencies of lines in 2 P 3/2  2 s 1/2 transition in Na Referring to diagram on previous page we have, from left to right