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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: K U E Kinetic energy Potential energy Total energy The kinetic energy K is conveniently given by K 1 2 mv 2 P 2 2m Where P is the momentum = m v (2) Schrödinger Equation Requirements (continued) 2. Consistency with de Broglie hypothesis P h and P k Where, λ and k are, respectively, the wavelength and the wave number. K (k ) 2m 2 (3 ) 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. ( x ) A sin( k x ) (4) To make sure that the solution is continuous, its derivative must have a value everywhere. d dx d A k cos( k x ) 2 dx 2 A k sin( k x ) k 2 k d 2 2 dx 2 2 (5 ) 2mK 2 2m 2 K (6) Time-Independent Schrödinger Equation d 2 dx 2 2 d 2m 2 (E U ) 2 2 m dx 2 U E (7 ) Probability, Normalization and Average • The probability density is given by 2 P ( x ) dx ( x ) dx 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 x2 P( x ) (x ) x1 2 dx By normalization we mean the total probability allover the space is 1, so that, (x ) 2 dx 1 The mean value ( the expectation value) of x, x p ( x ) x dx p ( x ) dx (x ) 2 x dx (x ) 2 dx If the wave function is normalized, therefore, x (x ) 2 x dx Applications • The Free Particle ( a particle is moving with no forces acting on it) • U(x)= constant anywhere=0 (arbitrarily) 2 d 2 m dx d d 2 2 2 E 2 dx 2 k 2 dx k 2 2 2 mE 2 2 mE 2 The solution of such deferential equation is given in the form Ψ(x)=A sin(kx)+B cos(kx) (8) Particle in a one dimensional box 0 (x ) A sin( k x ) B cos( k x ) at x 0 and at x L at 0 x L Finding A Ψ(0)=A sin(kx) + B cos(kx)=0 and for this, B=0 Therefore, ψ(x)=A sin(kx) (9) Ψ(L)=0, therefore A sin(kL)=0 since A≠0, sin(kL)=0 np kL=p, 2p, 3p, .. .., np (10) k L L 0 2 2 A sin ( np L 2n p sin( L 4 x) 1 Normalization condition np L x) x np L 2 2 A L 0 A 2 L The wave function is now given by, (x ) 2 L sin( np (11) x) L The energy is given by using (8) and (10) together, ( np L ) 2 2 mE 2 n p 2 En 2 2 mL 2 2 (12) The ground state (lowest ) energy EO is given by p 2 2 mL 2 2 EO We used n=1 The allowed energies for this particle are E n E O , 4 E O , 9 E O ,..... The wave function is shown here for n=1 and n=2 Ψ(x) n=1 x n=2 Example 5.2 P149 An 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 P151 Show 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). (x ) Since the wave function is 2 sin( np L x) L x And the average value is defined by L x [ 0 2 L sin( (x ) 2 x dx np 2 x )] x dx L L 2n p np x sin( x ) cos( 2 x) 2 2 x L L L x np np 2 L 4 2 4 8( ) L L 0 A particle in a two dimensional box The Schrödinger equation in two dimensions is 2 (x, y ) 2 (x, y ) U( x, y ) ( x, y ) E ( x, y ) 2 2 2m x y 2 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, ( x , y ) f ( x ) g( y ) f ( x ) A sin k x x B cos k x x g ( y ) A sin k y y B cos k y y 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 f ( x ) A sin k x x In the same way g ( y ) C sin k y y For x=L, f(x)=0 and y=L, g(y)=0 A sin k x L 0 This requires kx L=nxp with n=1,2,3 f ( x ) A sin nxp L x and g( y ) C sin nyp L y ( x , y ) f ( x ) g( y ) A ' sin nxp L x sin nyp y L To find the constant A’, the wave function should be normalized L L 0 ( A ' sin nxp L 0 x sin This integration gives ( x, y ) 2 L sin nxp L nyp L y ) dx dy 1 A' x sin 2 2 L nyp L y The energy states of a particle in a two dimensional box Substituting about the wave function ψ(x,y) in Schrödinger equation, we find 2 2p 2m L 3 2 nyp 2 nyp 2 pn x pn x 2 x ) sin( y ) (n x n y ) sin( x ) sin( y ) E sin( L L L L L Which after simplification becomes E 2 2p 3 2m L 2 (n n ) 2 x 2 y Chapter 7 The Hydrogen Atom Wave Functions • The Schrödinger Equation in Spherical Coordinates The Schrödinger equation in three dimensions is 2 2 2 2 U ( x , y, z ) E 2 2 2 2 m x y z The potential energy for the force between the nucleus and the electron is U 1 e 2 4 p O r 1 4 p O e 2 x y z 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. 2 2 2 Cartesian and spherical coordinates r cos θ z electron θ φ x r y x= r sin θ cos φ y= r sin θ sin φ z= r cos θ And Schrödinger equation becomes 2 2 1 1 2 sin 2 2 2 m r r r r sin r sin 2 U (r, , ) E 2 2 2 ( r , , ) This wave function can be written in terms of 3 functions in their corresponding variables, r, θ and φ ( r , , ) R ( r ) ( ) ( ) Hydrogen wave functions in spherical coordinates R(r) is called radial function Θ(θ ) is called polar function and Φ(φ) is called azimuthal function when 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 the principal quantum number l angular momentum quantum number ml magnetic quantum number n 1 2 2 2 1, 2, 3, … 0, 1, 2, ……±(n-1) 0, ±1, ±2, ……± l The energy levels of the hydrogen atom En me 4 1 32 p O n 2 2 2 2 The allowed values of the radius r around the nucleus are given by rn 4 p O me 2 2 n 2 Bohr radius (r at n=1) is denoted by aO and is given by aO 4 p O me 2 2 The Radial Probability Density P(r) The radial probability density of finding the electron at a given location is determined by P (r ) r R n , (r ) 2 2 The total probability of finding the electron anywhere around the nucleus is P (r ) dr 0 r 2 2 R n , (r ) dr 0 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 1 R 2 ,1 (r ) 3 (2a O ) r 3/2 e r 2aO aO 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. 1 r 2a P (r ) r e O 3/2 aO 3 ( 2 a ) O r 2 2 r d 2 1 r 2aO r e dr 3 ( 2 a O ) 3 / 2 a O r 3 r e 6a aO 5 O 4 r e 24 a r aO 6 O 0 Simplifying this result we get r 4 aO 2 0 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 P (r ) dr r R n , (r ) dr 0 0 2 2 r r aO 2 aO 2 r e 3 r 0 a2 o R 1,0 (r ) 2 3/2 e r / a O aO 2 2 2r 2r 2 dr 1 aO aO 2r aO e aO 0 0 . 323 32.3 % of the time the electron is closer than 1 Bohr radius to the nucleus. Angular Momentum 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 L r p 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 L ( 1) The z-components of L are given by L m 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 cos m ( 1) m ( 1) Intrinsic Spin iA i A pr q T T linear momentum q 2 Angular momentum vector 2pr v p mv rp 2m Using q=-e the charge of the electron, and rp= L , we get Magnetic moment due to electric current i L e L 2m 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 L ,z B e 2m Lz e 2m m Bohr Mgneton 2m L ,z m B B 9 . 274 X 10 e 24 J T Remember, ml =0, ±l An electric dipole in a uniform and non-uniform electric field A 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. Stern-Gerlach Experiment 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. 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 screen 2. 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 The magnitude of S, the spin angular momentum vector is given by Where s is the spin quantum number = ±½ S s ( s 1) Example 7.6 In a Stern – Gerlach type of experiment, the magnetic field varies with distance in the z dB T 1 . 4 direction according to dz The silver mm 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. z 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 U μ B zB z Problem 22 p 233 A 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 233 Consider 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