# The Normal Zeeman Effect

## Full text

(1)

7.1 Application of the Schrödinger Equation to the Hydrogen Atom

7.2 Solution of the Schrödinger Equation for Hydrogen

7.3 Quantum Numbers

7.4 Magnetic Effects on Atomic Spectra – Normal Zeeman Effect

7.5 Intrinsic Spin

7.6 Energy Levels and Electron Probabilities

### The Hydrogen Atom

By recognizing that the chemical atom is composed of single separable

electric quanta, humanity has taken a great step forward in the investigation of the natural world.

- Johannes Stark

(2)

The Dutch physicist Pieter Zeeman showed the spectral lines emitted by atoms in a magnetic field split into multiple energy levels.

 It is called the Zeeman effect.

If line is split into three lines ➝ Normal Zeeman effect

If line splits into more lines ➝ Anomalous Zeeman effect (Chapter 8)

Normal Zeeman effect (A spectral line is split into three lines):

Consider the atom to behave like a small magnet.

The current loop has a magnetic moment μ = IA and the period T = 2πr / v.

Think of an electron as an orbiting circular current loop of I = dq / dt

### 7.4: Magnetic Effects on Atomic Spectra The Normal Zeeman Effect

: Magnetic moment of Hydrogen atom

(3)

With no magnetic field to align them, point in random directions.

However in an external field the dipole will experience a torque which tends to align it in the field.

for any orientation the moment has a potential energy

Bohr magneton

(4)

### The Normal Zeeman Effect

Depending on the orientation of the magnetic moment (value of m), the potential of the orbiting electron in a magnetic field is:

Therefore each energy level (ℓ) of the orbiting electron which are (2 ℓ +1) degenerate will be split by the applied magnetic field.

 When a magnetic field is applied, the 2p level of atomic

hydrogen is split into three different energy states with energy difference of ∆E = μBB ∆m.

m Energy 1 E0 + μBB

0 E0

−1 E0 − μBB

(5)

### The Normal Zeeman Effect

A transition from 2p to 1s

3 different spectral lines due to photons emitted with slightly different

wavelengths from field split levels.

(6)

### Example 7.7

Determine the value of the Bohr magneton, and use this value to determine the energy difference between the m = 0 and +1 components of the 2p state of atomic hydrogen placed in a 2 Tesla magnetic field.

2

2 1/ 2 13.6 / 4 3.4

E  E   eV   eV

Relatively speaking E is small, however easily observed optically.

(7)

In 1920’s experiments were

performed to directly measure the space quantization of atoms.

In 1922 Stern and Gerlach

reported results of an experiment that clearly showed evidence for space quantization.

Experiment was based on the idea that an inhomogeneous magnetic field will exert a net force (in addition to a torque) on magnetic dipole.

(8)

### Stern Gerlach Experiment

Inhomogeneous magnetic field

If it moves through a homogeneous magnetic field, the forces exerted on opposite ends of the dipole cancel each other out and the trajectory of the

particle is unaffected.

(9)

An atomic beam of particles in the ℓ = 1 state pass through a magnetic field along the z direction.

The m = +1 state will be deflected down, the m = −1 state up, and the m = 0 state will be undeflected.

If the space quantization were due to the magnetic quantum number m, m states is always odd (2ℓ + 1) and should have produced an odd

number of lines. One single line (ℓ = 0), 3 lines (ℓ = 1), 5 lines (ℓ = 3), …

But, they observed only two lines!

The atoms are deflected either up or down!  Why?

(10)

### 7.5: Intrinsic Spin

Clearly there was a problem with number of lines observed in the Stern Gerlach experiment.

In 1925, Samuel Goudsmit and George Uhlenbeck in Holland proposed that the electron must have an intrinsic angular momentum and therefore a magnetic moment.

In order to explain experimental data, Goudsmit and Uhlenbeck proposed that the electron must have an intrinsic spin quantum number s = 1/2.

Paul Ehrenfest showed that the surface of the spinning electron should be moving faster than the speed of light!

Therefore this intrinsic spin angular momentum must be a purely quantum result.

### 

( 1) 3 S  s s  2

 

: spin angular momentum

(11)

s

### )

The spinning electron reacts similarly to the orbiting electron in a

magnetic field.  Spin will have quantities analogous to L, Lz, ℓ, and m.

like L, the electron’s spin can never be spinning with its magnetic moment μs exactly along the z axis.

The z – component of the spin angular momentum:

The magnetic spin quantum number ms has only two values, since (2s+1) = 2

depending on the value of ms , the spin will be either “up” or “down”.

Each electronic state in the atom now described by four quantum numbers (n, ℓ, m, ms)

s

sz



2 1 2 1 ms

(12)

### Intrinsic Spin

The magnetic moment due to intrinsic spin is

.

The coefficient of is −2μB

We can define the gyromagnetic ratio (for ℓ or s).

Where g = 1 and gs = 2, then

The z component of .

We are now in a position to understand the results of the Stern Gerlach experiment that produced only two distinct lines.

 If electrons are in an ℓ = 0 state no splitting due to m.

But, there is space quantization due to the intrinsic spin, ms.

B 2 e

m

, or 2 B

s s

e S S

m

        

2 , or

e B

L L

m

        

### 

 that of / is L  B

,s g ,sB L S,

    

2 ,

B B B B

s gs S S g L L

       

(13)

### Intrinsic Spin

So same argument that was used previously for the orbiting electron can now be applied the magnetic moment due to the intrinsic spin.

In a field the potential energy becomes:

Note: twice difference in ms levels than the difference in m levels.

Special Topic: 21 cm line transition (page 260)

Both proton and electron have intrinsic spin

lowest energy state is with them oppositely aligned.

∆E between opposite and parallel is 5.9 ×10-6 eV.

corresponds to a photon λ of 21 cm (radio wave)

widely used in Astronomy.

B s s 2 B

e e e

V B S B m B B B

m m m

 

             

2 2

B B

e e

V B L B m B m B

m m

 

          

, 2

B s B

VB

  

,

B B

VB

 

(14)

### 7.6: Energy Levels and Electron Probabilities

For hydrogen, the energy level depends on the principle quantum number n.

these energies are predicted with great accuracy by the Bohr model.

In a magnetic field the degeneracy is removed.

In many-electron atoms the degeneracy is also removed either because of internal B fields or because the average potential of an electron due to nucleus plus

electrons is non-Coulombic.

generally smaller ℓ states tends to lie at lower energy for a given n.

For example, sodium

E(4s) < E(4p) < E(4d) < E(4f).

(hydrogen atom)

(15)

### Selection rules for radiative transition

In ground state an atom cannot emit radiation.

It can absorb electromagnetic radiation, or gain energy through inelastic bombardment by

particles.

Can use the wave functions to calculate

transition probabilities for electrons changing from one state to another.

Allowed transitions: (for photon)

Electrons can absorb or emit photons to change states when ∆ℓ = ±1.

Change in ℓ implies a ∆LZ of ±ħ.

Conservation of angular momentum means photon takes up this angular momentum.

Forbidden transitions:(for photon)

Other transitions possible but occur with much smaller probabilities when ∆ℓ ≠ ±1.

1 0, 1 n anything

m

 

  

 

Selection rule

(16)

### (참고) Radiative transitions and Selection rules

- Concepts of Modern Physics, Beiser, Chapter 6.8 -

What happens when an electron goes from one state to another?

Energy level En 에서 Em 으로 떨어질 때의 복사선의 진동수 ()?

전자가 한 방향 (x-축 ) 으로 움직이는 system 을 고려하면 ,

양자 수 n인 (에너지 En 인 상태) 전자의 시간에 의존하는 파동함수 : Ψn

Ψn = (시간에 무관한 파동함수 n ) x (주파수 νn= En/h 인 시간 의존 함수)

이러한 전자의 위치에 대한 기대치 <x> 는

h E Emn

t h iE n

n

( n/ )

h E Emn

### 

 

dx e

x dx

x

x n n n n iEn / iEn / t

n

n

### 

Em En

h

(17)

 따라서, En 에너지 상태에 있는 전자의 <x> 는 n n* 위치만의 함수로 정의되기 때문에,

<x> 값은 시간에 따라 변하지 않는다.

 전자는 진동하지 않으므로 어떠한 복사(radiation)도 일어나지 않는다

 양자역학은 특정한 양자 상태에 머물러 있는 원자는 복사하지 않는다는 것을 예측하여 주고 있고 관측 결과와 일치한다.

Next, consider an electron that shifts from one energy state to another.

(다른 particle 과 충돌 혹은 radiation 을 흡수/방출하여 )

What is the frequency of radiation?

n 상태와 m 상태 모두 있을 수 있는 전자의 파동 함수 Ψ.

a*a  probability that the electron is in state n b*b  probability that the electron is in state m

n n



Em En

h

n n

## (참고)

(18)

전자가 위 상태 중 어느 곳에 있든지 복사 (radiation)을 발생하지 않음 .

그러나, m 에서 n 으로 천이하는 과정 (a 와 b 어느 것도 0이 아님) 에서는 전자기파가 발생된다.

중첩된 파동함수의 기대 값 <x>는

##   

*

n m n m

 

n n m n n m m m

### 

2 * * * * 2 *

a*a + b*b = a2 + b2 = 1

a= 1 , b = 0  e in the ground state (En) a= 0 , b = 1  e in the excited state (Em)

Em En

h

(iEn/ )h t n ne

 

n n

###     

 m me(iEm/ )h t

## (참고)

(19)

/   /

2 * * * iEm t iEn t

n n m n

 

###    

/   /

* * iEn t iEm t 2 *

n m m m

 

i

i

b a a b

dx

x E t

E

m n n

m n

m

* * * cos *

b a a b

dx

x E t

i Em n

m nn m

 

 

 sin ****

## (참고)

(20)

이 결과의 Real part 는 시간에 따라 다음과 같이 변한다.

따라서 전자는 삼각함수 모양으로 진동을 하게 되며, 그 진동수는

* 전자가 n 상태나 m 상태에 머물러 있을 때 전자의 위치에 대한 기대 값은 상수이다.

 진동하지 않고 복사도 안 일어난다

* 전자가 두 상태 사이에서 전이 할 때, 전자는 진동수 로 진동한다.

 이러한 전자는 electric dipole 같이 동일한 진동수 ( )를 갖는 electromagnetic wave 를 복사한다.

t h t

E t E

E

E m n cos 2

m n cos 2

cos  

 

 

 

 

 

h E Emn

Em En

h

## (참고)

(21)

Some transitions are more likely to occur than others  Selection rule

진동수  를 알기 위해서는 시간의 함수로서 a, b 의 확률이나 파동함수n과 m를 알 필요는 없다.

그러나 , 어떤 전이가 일어날 확률을 계산하기 위해서는 이 값들이 필요하다.

여기 상태에 있는 원자가 복사하기 위해 필요한 일반적인 조건은 적분

이어야 하는데, 그것은 복사 강도가 적분 값에 비례하기 때문이다.

허용된 전이 (allowed transition) ← 적분 값 ≠0 인 경우 금지된 전이(forbidden transition) → 적분 값 =0 인 경우

*  0

xnm dx

## (참고)

(22)

수소 원자의 경우 복사 전이에 관여하는 처음상태와 나중상태를 규정하기 위해three quantum numbers 가 필요하다.

Initial state : n’, l’, ml Final state : n, l, ml

수소원자의 n,l,ml을 알고 있으므로 u=x, u=y, u=z 인 경우 위의 값을 구할 수 있다.

그 결과 l 이 +1과 –1 만큼 변하고 ml 이 바뀌지 않거나 +1 또는 –1 만큼 바뀌는 전위만 일어난다.

주 양자 수n의 변화는 제한 받지 않는다.

## 

, , * ,' ,' l'

l n l m

m l

n

### 

선택 규칙l = +- 1 에 의해 허용되는 전이를 보여주는 수소 원자의 에너지 준위 그림.

Allowed  transition

Selection rules

l = ±1

ml = 0, ±1

 원자가 복사를 일으키기 위해서l =  1 이 되어야 한다는 selection rule 은 photon 이 처음과 나중의 각 운동량 차이에 해당하는 각 운동량(angular momentum)  h 를 갖고 방출된다는 것이다.

 각운동량  h 를 가지고 있는 광자란 좌원편광 혹은 우원 편광된 전자기파라는 의미임.

## (참고)

(23)

### (참고) Normal Zeeman effect and Selection rules

- Concepts of Modern Pgysics, Chapter 6.10, Beiser-

자기장 내에서는 특정 원자 상태의 에너지는 n 값 뿐만 아니라 ml 의 값에도 관계한다.

복사 전이의 경우,

 ml = 0, 1 로 제한되기 때문에 서로 다른 l 를 가진 갖는 두 상태에서 생기는 spectrum 선은 3개의 진동수만 갖게된다.

진동수가 0 spectrum 선이 다음과 같은 진동수를 갖는 3개의 성분으로 분리됨.

m B e h

B

m B e h

B

B B

4 4

0 0

3

0 2

0 0

1

Normal Zeeman effect

(24)

### Probability Distribution Functions

Unlike the planetary model of Bohr, in Schrödinger wave picture the “position” of the electron in each state, defined by 4

quantum numbers (n, ℓ, m, ms), is spread over space and is not well defined.

We must use the wave functions to calculate the probability distributions of the electrons

The probability of finding the electron in a differential volume element d is:

in spherical polar coordinates:

ϕ dependence:

Ig(ϕ)I2 contains factor (eimℓϕ)* eimℓϕ = 1 thus no ϕ dependence in probability  thus wave functions are symmetric about z-axis.

(25)

### Probability Distribution Functions

If we are only interested in the radial probability distributions of the electron,

and, if the functions f(ϴ) and g(ϕ) have been properly normalized then integration over all values of ϴ, and ϕ yields:

Therefore,

Note: the radial probability density P(r) = r2|R(r)|2 depends only on n and l.

(26)

Observations:

< r > depends on n or energy, and is

independent of ℓ.

< r > ∝ n2

(same as Bohr).

most probable radii for 1s and 2p agree with Bohr result (a0 and 4a0 respectively).

true for wave functions with maximum ℓ

n

n

### ( )  r R

2 n 2

R(r) and P(r) for the lowest-lying states of the hydrogen atom

(27)

n

n

### ( )  r R

2 n 2

(28)

Example 7.11

Find the most probable radius for the electron in a hydrogen atom in the 1s state.

To find the maximum we set the derivative of the radial probability equal to zero.

Example 7.12: Calculate the average orbital radius of 1s.  < r > = (3/2) a0

n

### ( )r R

2 n 2

(29)

Example 7.13

What is the probability of an electron in the 1s state of the hydrogen atom being at a radius greater than the Bohr radius?

n

2 n 2

(30)

(31)

n m

Updating...

## References

Related subjects :