Modern Physics 342
References :1. Modern Physics by Kenneth S. Krane. , 2nd Ed.
John Wiley & Sons, Inc.2. Concepts of Modern Physics by A. Beiser, 6th Ed.
(2002), McGraw Hill Com.3. Modern Physics for Scientists and Engineers by
J. Taylor, C. Zafiratos and M. Dubson, 2nd Ed, 2003.
Chapters: 5 (Revision), 6 (6.4), 7, 8, 10, 11 and 12
Ch. 5 (Revision)
Schrödinger Equation
Schrödinger Equation Requirements
1. Conservation of energy is necessary:
EUK
Kinetic energyPotential energy
Total energy
The kinetic energy K is conveniently given by
)2(m2
Pmv21K
22
Where P is the momentum = m v
Schrödinger Equation Requirements (continued)
2. Consistency with de Broglie hypothesis
kPandhP
Where, λ and k are, respectively, the wavelength and the wave number.
)3(m2)k(K
2
Schrödinger Equation Requirements (continued)
3. Validity of the equation
The solution of this equation must be valid everywhere, single valued and linear.
By linear we meant that the equation must allow de Broglie waves to superimpose properly.
The following is a mathematical form of the wave associating the particle.
)4()xksin(A)x(
To make sure that the solution is continuous, its derivative must have a value everywhere.
)xkcos(kAdxd
)5(k)xksin(kAdxd 22
2
2
22 Km2k
)6(Km2dxd
22
2
Time-Independent Schrödinger Equation
)UE(m2
dxd
22
2
)7(EUdxd
m2 2
22
Probability, Normalization and Average
• The probability density is given by
Which is the probability of finding a particle in the space dx.
The probability of finding this particle in a region between x1 and x2 is
dx)x(dx)x(P 2
dx)x()x(P2
1
x
x
2
By normalization we mean the total probability allover the space is 1, so that,
1dx)x( 2
dx)x(
dxx)x(
dx)x(p
dxx)x(px
2
2
dxx)x(x 2
The mean value ( the expectation value) of x,
If the wave function is normalized, therefore,
Applications
• The Free Particle ( a particle is moving with no forces acting on it)• U(x)= constant anywhere=0 (arbitrarily)
The solution of such deferential equation is given in the form
Ψ(x)=A sin(kx)+B cos(kx)
Edxd
m2 2
22
22
2 mE2dxd
22
2
kdxd
22 mE2k
(8)
Particle in a one dimensional box
Lx0at)xkcos(B)xksin(A
Lxatand0xat0)x(
Finding A
Ψ(0)=A sin(kx) + B cos(kx)=0 and for this, B=0Therefore,
ψ(x)=A sin(kx) (9)
Ψ(L)=0, thereforeA sin(kL)=0 since A≠0, sin(kL)=0kL=p, 2p, 3p, .. .., np
Lnk p
1)xLn(sinA
L
0
22 p
LAn
2
xLn
4
)xLn2sin(
2
L
0
p
p
p
L2A
Normalization condition
(10)
The wave function is now given by,
The energy is given by using (8) and (10) together,
)xLnsin(
L2)x( p
(11)
22 mE2)
Ln(
p2
222
n mL2nE p
(12)
The ground state (lowest ) energy EO is given by
2
22
O mL2E p
We used n=1
The allowed energies for this particle are
,.....E9,E4,EE OOOn
The wave function is shown here for n=1 and n=2
Ψ(x)
x
n=1
n=2
Example 5.2 P149An electron is trapped in a one-dimensional region of length 1X10-10 m. How much energy must be supplied to excite the electron from the ground state to the first excited state? In the ground state, what is the probability of finding the electron in the region from 0.09 X 10-10 m to 0.11 X 10-10 m? In the first excited state, what is the probability of finding the electron between x=0 and x=0.25 X 10-10 m?
Example 5.3 P151Show that the average value of x is L/2, for a particle in a box of length L, independent of the quantum state (not quantized).
)xLnsin(
L2)x( p
Since the wave function is
And the average value is defined by
dxx)x(x 2
p
L
0
2 dxx)]xLnsin(
L2[x
2L
4x
)Ln(8
)xLn2cos(
Ln4
)xLn2sin(x
L2x
L
0
2
2
p
p
p
p
A particle in a two dimensional box
The Schrödinger equation in two dimensions is
)y,x(E)y,x()y,x(Uy
)y,x(x
)y,x(m2 2
2
2
22
U(x,y)=0 inside the box (0≤x≤L) & (0≤y≤L)
U(x,y)=∞ outside the box
The wave function ψ(x,y) is written as a product of two functions in x and y,
)y(g)x(f)y,x(
xkcosBxksinA)x(f xx
ykcosByksinA)y(g yy Since ψ(x,y) must be zero at the boundaries,
ψ(0,y) =0 ψ(L,y) =0 ψ(x,0) =0 ψ(x,L) =0
Therefore, A sin kx (0)+ B cos kx (0)=0 which requires B=0
xksinA)x(f xIn the same way
yksinC)y(g y
For x=L, f(x)=0 and y=L, g(y)=0
0LksinA x This requires kx L=nxp with n=1,2,3
xL
nsinA)x(f xp and yL
nsinC)y(g yp
To find the constant A’, the wave function should be normalized
yL
nsinx
Lnsin'A)y(g)x(f)y,x( yx p
p
1dydx)yL
nsinx
Lnsin'A( 2yx
L
0
L
0
p
p
This integration gives L2'A
yL
nsinx
Lnsin
L2)y,x( yx p
p
The energy states of a particle in a two dimensional box
Substituting about the wave function ψ(x,y) in Schrödinger equation, we find
E)yL
nsin()x
Lnsin(
L2)nn()y
Ln
sin()xLnsin(
L2
m2yx2
y2x
yx3
22
p
p
p
pp
Which after simplification becomes
)nn(L
2m2
E 2y
2x3
22
p
Chapter 7The Hydrogen Atom Wave Functions
• The Schrödinger Equation in Spherical CoordinatesThe Schrödinger equation in three dimensions is
The potential energy for the force between the nucleus and the electron is
This form does not allow to separate wave function Ψ into functions in terms of x, y and z, so we have to express the whole equation of Schrödinger in terms of spherical coordinates, r, θ, and φ.
E)z,y,x(U
zyxm2 2
2
2
2
2
22
222
2
O
2
O zyxe
41
re
41U
p
p
Cartesian and spherical coordinates
θ
φ
r
x
y
z
electron
r sin θ
r sin θ co
s φ
r cos
θ
r sin θ sin φ
x= r sin θ cos φy= r sin θ sin φz= r cos θAnd Schrödinger equation becomes
E),,r(Usinr1sin
sinr1
rr2
rm2 2
2
2222
22
),,r(
)()()r(R),,r(
This wave function can be written in terms of 3 functions in their corresponding variables, r, θ and φ
Hydrogen wave functions in spherical coordinates
R(r) is called radial functionΘ(θ ) is called polar function and Φ(φ) is called azimuthal functionwhen solving the three differential equations in R(r), Θ(θ ) and Φ(φ), l and ml quantum numbers were obtained in addition to the previous principal quantum number n obtained before.
n
1
2
2
2
n the principal quantum number 1, 2, 3, …l angular momentum quantum number 0, 1, 2, ……±(n-1)ml magnetic quantum number 0, ±1, ±2, ……± l
The energy levels of the hydrogen atom
222O
2
4
n n1
32emE
p
The allowed values of the radius r around the nucleus are given by
22
2O
n nme
4r p
2
2O
O me4a p
Bohr radius (r at n=1) is denoted by aO and is given by
The Radial Probability Density P(r)
The radial probability density of finding the electron at a given location is determined by
2,n
2 )r(Rr)r(P
The total probability of finding the electron anywhere around the nucleus is
0 0
2,n
2 dr)r(Rrdr)r(P
The limits of the integration depend on the conditions of the problem
Example 7.1 Prove that the most likely distance from the origin of an electron in the n=2, l=1 state is 4aO .
At n=2 and l =1, R2,1 (r) is given by
Oa2r
O2/3
O1,2 e
ar
)a2(31)r(R
The most likely distance means the most probable position. The maximum value of the probability is obtained if r=4aO .To prove that, the first derivative of P(r) with respect to r is zero at this value.
2
a2r
O2/3
O
2 Oear
)a2(31r)r(P
0ear
)a2(31r
drd
2
a2r
O2/3
O
2 O
0a24
era6er
6O
ar
4
5O
ar
3 OO
Simplifying this result we get
Oa4r
Example 7.2 An electron in the n=1, l=0 state. What is the probability of finding the electron closer to the nucleus than the Bohr radius aO ?
The probability is given by
0 0
2,n
2 dr)r(Rrdr)r(P Oa/r
2/3O
0,1 ea
2)r(R
323.0
ear2
ar21dre
a
2rO
O
OO
ar
0r
a
0
ar2
2O
2
O
2
ar
23
o
2
32.3 % of the time the electron is closer than 1 Bohr radius to the nucleus.
Angular Momentum
)1(L
mL
We discussed the radial part R(r) of Schrodinger equation. In this section we will discuss the angular parts of the Schrodinger equation.
The classical angular momentum vector is given by prL
During the variables separation of wave functions in Schrodinger equation, angular momentum quantum number l was produced. The length of the angular momentum vector L is given by
The z-components of L are given by
where ml is the magnetic quantum number 0, ±l
The angular momentum vector components
For l=2, ml =0, ±1, ±2
The angle is given by
)1(m
)1(mcos
Intrinsic Spin
Angular momentum vector
Magnetic moment due to electric current i
Ai 2rA p
Tqi
vr2T p
mvpmomentumlinear
prm2q
Using q=-e the charge of the electron, and rp= L , we get
Lm2e
L
The negative singe indicates that µL and L work in opposite directions.
When the angular momentum vector L is inclined to the direction of the z-axis, the magnetic moment µL has a z-component given by
mm2eL
m2e
zz,L
MgnetonBohrm2
eB
Bz,L m
TJ10X274.9 24
B
Remember, ml =0, ±l
An electric dipole in a uniform and non-uniform electric fieldA magnetic dipole in a non-uniform magnetic field
The electric dipole has its moment p rotates to align with the direction of the electric field
Two opposite dipoles in the same non-uniform electric field are affected by opposite net forces that lead to displacing each dipole up and down according to their respective alignments.
Similarly, the magnetic dipoles are affected in the same way.
When an electron with an angular momentum inclined to the magnetic filed, it may move up or down according to the direction of rotation around the nucleus.
A beam of hydrogen atoms is in the n=2, l= 1 state. The beam contains equal numbers of atoms in the ml = -1, 0, and +1 states. When the beam passes a region of non-uniform magnetic field, the atoms with ml =+1 experience a net upward force and are deflected upward, the atoms with ml =-1 are deflected downward, while the atoms with ml =0 are undeflected.
Stern-Gerlach Experiment
After passing through the field, the beam strikes a screen where it makes a visible image. 1. When the filed is off, we expect to see one image of the slit in the center of
the screen2. When the field is on, three images of the slit on the screen were expected –
one in the center, one above the center (ml =+1 ) and one below (ml =-1). The number of images is the number of ml values = 2l+1= 3 in our example.
• In the Stern - Gerlach experiment, a beam, of silver atoms is used instead of hydrogen.
• While the field is off, and instead of observing a single image of the slit, they observed two separate images.
The experiment
)1s(s SThe magnitude of S, the spin angular momentum vector is given by
Where s is the spin quantum number = ±½
Example 7.6
In a Stern – Gerlach type of experiment, the magnetic field varies with distance in the z direction according to The silver atoms travel a distance x=3.5 cm through the magnet. The most probable speed of the atoms emerging from the oven is v=750 m/s. Find the separation of the two beams as they leave the magnet. The mass of a silver atom is 1.8 X 10-25 kg, and its magnetic moment is about 1 Bohr magneton.
mmT4.1
dzdBz
3.5 cm
vO =750 m/s. The force applied to the beam must be obtained
The force is the change of potential energy U with distance z. The potential energy U is given by
zzBU Bμ
Problem 22 p 233A hydrogen atom is in an excited 5g state, from which it makes a series of transitions, ending in the 1s state. Show on an energy levels diagram the sequence of transitions that can occur. Repeat the last steps if the atom begins in the 5d state.
Problem 23 p 233Consider the normal Zeeman effect applied to 3d to 2p transition. (a) sketch an energy-level diagram that shows the splitting of the 3d and 2p levels in an external magnetic field. Indicate all possible transitions from each ml state of the 3d level to each ml state of the 2p level. (b) which transitions satisfy the Dml =±1 or 0 selection rule? (c) that there are only three different transitions energies emitted.
For ℓ=2, mℓ=+2,+1,0,-1,-2
Top Related