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ELEKTRONIKOS PAGRINDAI 1 2008 1. PROPERTIES OF ELECTRONS 1. Electrons are microparticles. 2. Quantum mechanics allows to reveal properties of microparticles. 3. Quantum theory, the branch of physics which is based on quantization, began in 1900 when Karl Ernst Ludwig Planck (1858 – 1947) published his theory explaining the emission spectrum of black bodies. K. Planck stated that … electromagnetic energy could be emitted only in quantized form, … the energy could only be a multiple of an elementary unit W = hν. 4. Quantum mechanics was developed by N. Bohr, W. Heisenberg, P. Dirac, E. Schrődinger, V. Pauli and other famous physicists. Objectives: Gain knowledge about properties of electrons (using principles of quantum mechanics), that is necessary for understanding of semiconductor physics and studies of semiconductor and electronic devices. VGTU EF ESK [email protected] ELEKTRONIKOS PAGRINDAI 2 2008 Wave-particle duality of microparticles. De Brouglie waves Light has dual nature: λ = c /ν W = hν = hω W = mc 2 p = mc = W / c = hν / c = h / λ 1924: L. de Brouglie postulated: The wave-particle duality is the common property of microparticles. The length and the frequency of the wave, that correspond to the microparticle with energy W and momentum p, are given by ν =W /h λ =h/ p Equations are called de Broglie’s equations. Experiments made by K. Davison and L. Gemer in 1927 approved the theory about the wave-particle duality . De Broglie’s waves are not electromagnetic waves. They are probability waves. VGTU EF ESK [email protected] ELEKTRONIKOS PAGRINDAI 3 2008 The uncertainty principle From classical mechanics we know that six parameters define the state of a particle: three co-ordinates and three projections of momentum. If we know the momentum of a microparticle, we can find the length of the wave that corresponds to this microparticle. It means that a monochromatic harmonic wave corresponds to the microparticle. This wave exists in the space from –∞ to +∞. So, if we know the momentum p, we cannot find the co-ordinate of the particle: the lack of precision of the co-ordinate of the particle is infinite. In 1927 Heisenberg published the uncertainty principle. He showed that, if an attempt is made to measure certain pairs of variables concerned with a physical system, then the lack of precision with which the two variables can be specified simultaneously becomes apparent. ∆x∆p ≥ h ∆x∆px ≥ h, ∆y∆p y ≥ h, ∆z∆p z ≥ h. ...we cannot measure co-ordinate and momentum simultaneously with unlimited accuracy. The inequalities represent fundamental limitation imposed by nature. VGTU EF ESK [email protected] ELEKTRONIKOS PAGRINDAI 4 2008 The Schrodinger wave equation According to Heizenberg’s inequalities we cannot find the co-ordinates and momentum of a microparticle simultaneously. So we cannot apply Newton’s laws to a microparticle. According to de Broglie microparticles behave in a wave-like manner. Waves are defined by wavefunctions. In 1926 Schrodinger proposed an equation for predicting the wavefunction of a microparticle. Now the theory based on Schrodinger equation is called wave or quantum mechanics. The wavefunction can be found from the equation: ∂Ψ (x, y, z , t ) h 2 − jh = ∆Ψ (x, y, z , t ) − Wp (x, y, z , t ) Ψ (x, y, z , t ) ∂t 2m Ψ ( x, y, z, t ) = ψ ( x, y, z )ϕ (t ) VGTU EF ESK [email protected] ELEKTRONIKOS PAGRINDAI 5 2008 The Schrodinger wave equation and its solution ∂Ψ (x, t ) h 2 − jh = ∆Ψ (x, t ) − Wp (x ) Ψ (x, t ) ∂t 2m Ψ ( x, t ) = ψ ( x)ϕ (t ) 2 h 2 1 d ψ ( x) 1 d ϕ (t ) − Wp = − jh . 2 2m ψ ( x ) d x ϕ (t ) d t The equation can be simplified by separating out the time-depending and position-depending parts: d 2 ψ ( x) dx 2 d ϕ (t ) dt + 2m h = −j 2 [W − Wp ( x)]ψ ( x) = 0. W ϕ (t ). h ϕ (t ) = C exp − j W t = C e − jωt h ... if the energy is constant, the wavefunction is a harmonic function of time t. VGTU EF ESK [email protected] ELEKTRONIKOS PAGRINDAI 6 2008 Properties of a free microparticle A microparticle is free, if no force acts it and its potential energy Wp = 0. Then W = Wk, and the Schrodinger equation can be simplified: d 2 ψ ( x) dx 2 d 2 ψ ( x) dx 2 + 2m h 2 Wk ψ ( x ) = 0 + k ψ ( x) = 0 2 mv 2 p2 h2 Wk = = = 2 2m 2mλ2 k = 2π / λ = p / h = ω / c ψ ( x) = ψ 1 ( x ) + ψ 2 ( x) = A1 e jkx + A 2 e − jkx k is the propagation constant or wave number of the de Broglie wave, corresponding to the microparticle. The solution describes the superposition of two waves. The first term represents the incident probability wave and the second term represents the reflected wave travelling in the opposite direction. h 2 4 π 2 h 2 2 ... the kinetic energy of a free microparticle may have Wk = = k any value. 2 2m λ 2m …the energy spectrum of the free microparticle is continuous. VGTU EF ESK [email protected] ELEKTRONIKOS PAGRINDAI 7 2008 Conclusions 1. The wave-particle duality is the common property of microparticles. 2. The length and the frequency of the wave, that correspond to the microparticle with energy W and momentum p, are given by ν =W /h λ =h/ p 3. De Broglie’s waves are not electromagnetic waves. They are probability waves. 4. It is not possible to measure the co-ordinate and momentum of a microparticle simultaneously with unlimited accuracy. According to the Heisenberg’s uncertainty principle ∆x∆p ≥ h VGTU EF ESK [email protected] ELEKTRONIKOS PAGRINDAI 8 2008 Conclusions 5. According to de Broglie, microparticles behave in a wave-like manner. Waves are defined by wavefunctions. The wavefunction, describing a micropartcle, can be found as a result of solving Schrodinger equation. 6. When potencial energy is constant (does not depend on time), the Schrodinger equation can be simplified by separating out the time-depending and position-depending parts. 7. A microparticle is free, if no force acts it and its potential energy is 0. The energy spectrum of the free microparticle is continuous. VGTU EF ESK [email protected] ELEKTRONIKOS PAGRINDAI 9 2008 Properties of a free microparticle. Problems 1. The length of de Broglie’s wave that corresponds to the electron is 0.1 nm. Find the frequency of the wave. 2. Prove that Ψ·Ψ*=|Ψ|2. 3. How are the phase velocity, and the group velocity related to the velocity of a free microparticle? VGTU EF ESK [email protected] ELEKTRONIKOS PAGRINDAI 10 2008 Microparticles and potential barriers Let us consider what happens when a microparticle impinges on a potential barrier. Steps of solution of the problem: 1. Creation of the model. 2. Solution of the Schrodinger equation. General solution is found. 3. Application of boundary conditions. 4. Finding of reflection and transmission factors, describing probabilities of reflection and passing the barrier by the particle. 5. Analysis of derived expressions in order to find: • what happens when a microparticle meets a low barrier; • what happens when a microparticle meets a high barrier. VGTU EF ESK [email protected] ELEKTRONIKOS PAGRINDAI 11 2008 Microparticles and potential barriers In order to reveal what happens when a microparticle meets the barrier, we must solve the Schrodinger equation. 1 region d 2 ψ 1 ( x) 2 region d x2 + 2m Wψ 1 ( x ) = 0 2 h ψ 1 ( x) = A1 e jk1 x + B1 e − jk1 x k1 = d 2 ψ 2 ( x) d x2 k2 = 2π λ2 VGTU EF ESK + = 2m (W − Wb )ψ 2 ( x) = 0 2 h 2m(W − Wb ) h 2π λ1 2mW h = ψ 2 ( x ) = A 2 e jk 2 x + B 2 e − jk 2 x B1 A1 2 2 2 =R [email protected] λ1 A2 =D 2 λ 2 A1 R + D =1 ELEKTRONIKOS PAGRINDAI 12 2008 Microparticles and potential barriers ... Boundary conditions must be satisfied: ψ 1 (0) = ψ 2 (0), Reflection coefficient ψ 1 ' (0) = ψ 2 ' (0). k1 − k 2 R = k1 + k 2 2 •... a microparticle can be reflected by the low potential barrier. • If the height of the low potential barrier increases, the reflection coefficient also increases and becomes 1 when the height of the low barrier becomes equal to the energy of the microparticle. If the barrier becomes high, a microparticle is reflected. k2 = jk 2∗ =j 2m(Wp − W ) h 2 2 ψ 2 = A2 e − 2 k 2• x •... microparticles penetrate into the high potential barrier before reflection from the barrier. • if the barrier is sufficiently thin, there is a small but finite probability that a microparticle can penetrate the barrier. • This is the essence of the tunnel effect. VGTU EF ESK [email protected] ELEKTRONIKOS PAGRINDAI 13 2008 A microparticle in a one dimensional potential well Let us reveal properties of a microparticle being in a potential well. Steps of solution of the problem: 1. Creation of the model taking into account practically important cases. 2. Solution of the Schrodinger equation. General solution is found. 3. Application of boundary conditions. 4. Analysis of derived expressions in order to find: • energy spectrum of a microparticle; • probability to find microparticle. VGTU EF ESK [email protected] ELEKTRONIKOS PAGRINDAI 14 2008 A microparticle in a one dimensional potential well In order to reveal properties of a microparticle, we must solve the Schrodinger equation. d 2 ψ ( x) d x2 k= 2π λ 2m + 2 Wkψ ( x) = 0 h = 2mW h ψ ( x) = A e jkx + B e − jkx ψ (0) = 0, ψ (a) = 0. ψ ( x) = 2 j A sin kx = C sin kx ψ (a) = C sin ka = 0 nπ k= a h2 2 h2 2 W= k = n 2 2m 8ma n is integer (n =1, 2, 3,…). It is called the quantum number. ... The energy spectrum of a microparticle in a potential well is discrete. The total energy of a microparticle has particular allowed values that are dependent upon quantum number n. VGTU EF ESK [email protected] ELEKTRONIKOS PAGRINDAI 15 2008 Conclusions 1. A microparticle can be reflected by the low potential barrier. If the height of the low potential barrier increases, the reflection coefficient also increases and becomes 1 when the height of the low barrier becomes equal to the energy of the microparticle. 2. Microparticles penetrate into the high potential barrier before reflection from the barrier. If the barrier is sufficiently thin, there is a small but finite probability that a microparticle can penetrate through the barrier. This is the essence of the tunnel effect. 3. The energy spectrum of a microparticle in a potential well is discrete. The total energy of a microparticle has particular allowed values that are dependent upon quantum number n. VGTU EF ESK [email protected] ELEKTRONIKOS PAGRINDAI 16 2008 Properties of microparticles. Problems 1. An electron having kinetic energy of 10 eV meets the potential barrier that is 6 eV high. Find probability that electron will be reflected by the barrier. 2. Compute the three lowest energy levels for an electron trapped in a onedimensional potential well of length 0.1 nm. 3. An electron is confined in a one-dimensional potential energy well of length 0.3 nm. Find the kinetic energy of the electron when in the ground state and the spectral frequency resulting from a transition from the next state (n = 2) to the ground level. 4. An electron is in the infinitively deep one-dimensional potential well having rectangular form. The wave function ψ(x)=Csin(nπx/a) corresponds to the electron. Using normalization condition find constant C at n = 1. VGTU EF ESK [email protected] ELEKTRONIKOS PAGRINDAI 17 2008 Electrons in atoms Taking into account properties of a microparticle in a potential well, let us consider states of an electron in a hydrogen atom and states of electrons in many-electron atoms. A hydrogen-like atom consists of a central nucleus surrounded by an electron. The potential energy of the electron is given by Zq Zq 2 Wp = − 4πε 0 r According to the graph the electron is trapped in the potential well. The well is three-dimensional. VGTU EF ESK [email protected] ELEKTRONIKOS PAGRINDAI 18 2008 Electrons in atoms Elektronų savybės In order to reveal properties of the electron it is necessary to solve the threedimensional Schrodinger equation using spherical coordinate system. Because the well is three dimensional, the solution is more complicate. Let us skip the solution and turn our attention to the main results. Because the electron in a hydrogenic atom moves in three-dimensional space and the Schrodinger equation is also three-dimensional, three quantum numbers are necessary to characterise the state of the electron in the atom. The first quantum number is the main or principal quantum number. It may have values : n =1; 2; … It defines the total energy of the electron: Z 2 mq 4 1 Rch Wn = − 2 2 2 = − Z 2 2 8ε 0 h n n ... The electron can occupy allowed energy levels. At n = 1, the atom and the electron are in the ground state. VGTU EF ESK Z = 1, W1 = –13,6 eV [email protected] ELEKTRONIKOS PAGRINDAI 19 2008 Electrons in atoms The second quantum number is called the orbital or azimuthal quantum number: l = 0; 1; ... ; (n-1). l = 0 – s-state; l = 1 – p- state; l = 2 – d- state; l = 3 – f- state, … The second quantum number is associated with the angular momentum of the electron which is itself quantized. L = mr × v L = h l (l + 1 ) The third quantum number is called the magnetic quantum number. It can take values: –l, ..., –1, 0; 1;..., l . The values of the magnetic quantum number are associated with the projection of the angular momentum on the direction of the external magnetic field. L =mh H l In 1922 O. Stern and W. Gerlach found that … if the beam from the electron gun is directed into inhomogeneous magnetic field, the beam is split into two parts. http://www.upscale.utoronto.ca/GeneralInterest/Harrison/SternGerlach/SternGerlach.html http://en.wikipedia.org/wiki/Stern-Gerlach_experiment VGTU EF ESK [email protected] ELEKTRONIKOS PAGRINDAI 20 2008 Electrons in atoms In 1925 R. Kronig, G. Uhlenbeck and S. Goudsmit proposed the concept of elementary particle spin. ... a further quantum number is necessary to define completely a particular quantum state which is not apparent from the solution of the three-dimensional Schrodinger equation. It is called the spin quantum number. Its possible values 1/2 and –1/2 correspond to the spin of the electron in an either clockwise or anticlockwise direction.: LsH = ms h … four quantum numbers are necessary to define the quantum state of the electron in the hydrogen atom or hydrogenic atom. Solving of the Schrodinger equation in the case of a many-electron atom gives that the state of an electron is also defined by four quantum numbers. The total energy of an electron in the many-electron atom depends on the values of the first and second quantum numbers, n and l . VGTU EF ESK [email protected] ELEKTRONIKOS PAGRINDAI 21 2008 Electrons in many-electron atoms Besides nucleus other electrons of a manyelectron atom act on an electron. As a result there are more allowed energy levels in the manyelectron atom, corresponding to quantum numbers n and l. Electrons that have the same principal quantum number are said to be in the same shell. A shell may contain subshells that correspond to values of the azimuthal quantum number. The states that correspond to subshells usually are denoted using designation letters s, p, d, f that correspond to values 0, 1, 2, 3 of the orbital quantum number. VGTU EF ESK [email protected] ELEKTRONIKOS PAGRINDAI 22 2008 Electrons in many-electron atoms Electrons tend to fill the lowest allowed energy levels first. According to Pauli’s exclusion principle, no more than one electron can exist in any one quantum state. Electronic structures of various chemical elements may be found and the structure of the periodic table may be constructed on the basis of these two principles and the quantum mechanical theory of atoms. VGTU EF ESK [email protected] ELEKTRONIKOS PAGRINDAI 23 2008 Properties of electrons. Problems 1. How many electrons may be in the states corresponding to the various sets of quantum numbers? 2. Find the electronic structures of Si and Ge atoms. 3. A hydrogen atom is excited from a state with n =1 to one with n = 4. Calculate and display on the energy-level diagram the different photon energies that may be emitted if the atom returns to its ground state. 4. The helium ion He+ has emitted energy quantum of 7.5 eV. What happened in the ion? 5. A photon with a wavelength of 75 nm knocks out an electron from the hydrogen atom. Find the length of the de Broglie wave corresponding to the electron that becomes free. VGTU EF ESK [email protected]