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Translational Motion - Free Motion in One Dimension • A particle of mass m free to move in one dimension ( along entire x-axis) with zero potential energy • Ek = k2ћ2/2m – Ek is the kinetic energy determined from the Schrödinger equation – ћ = h/2 • All values of the energy, Ek, are permitted – k can take on any value for a particle free to move along entire x-axis • The translational energy of a free particle is not quantized Chapter 8 1 A Particle in a Box • A particle of mass m confined between two walls at x=0 and x=L • The potential energy is zero inside the box but rises to infinity at the walls – an infinite square well • k(x) = C coskx + D sinkx – This is the general solution of the Schrödinger equation for the particle inside the box – Ek = k2ћ2/2m – ћ = h/2 Chapter 8 2 Particle in a Box – Acceptable Solutions • There is no chance of finding the particle outside the box. – is zero outside the box • The amplitude of wavefunction must be zero at each end of the box ( must be continuous) • Boundary conditions: k(0) = 0 and k(L)=0 • k(0) = C – sin0 =0 and cos0 =1 • Thus, C must be zero • k = D sinkx • k(L) = D sinkL = 0 • sinkL must be zero – D cannot be zero (this would mean that =0 for all x-values which is impossible) Chapter 8 3 Particle in a Box – Possible Wavefunctions • sin kL = 0 • kL = n n = 1, 2, 3, …….. (any integer) – n=0 is ruled out because k would be zero • k = n/L • n(x) = D sin (nx/L) n = 1, 2, 3, …. – These are the possible wavefunctions • En = n2h2/8mL2 n = 1, 2, 3, …… • The energy is quantized • To satisfy the boundary conditions, only certain wavefunctions are possible – Each wavefunction corresponds to a certain energy – The energy can only have discrete values Chapter 8 4 Particle in a Box – Normalization of the Wavefunctions • n(x) = D sin (nx/L) • * dx = 1 – Integrated from x=0 to x=L – The particle must be found inside the box • D = (2/L)½ • n(x) = (2/L)½ sin (nx/L) • En = n2h2/8mL2 n = 1, 2, 3, …….. • A quantum number, n, is used to label energies and wavefunctions – A quantum number is an integer (or in some cases a half-integer) that labels the state of the system Chapter 8 5 Particle in a Box – Properties of the Wavefunctions • All wavefunctions for the particle in the box have the same amplitude but different wavelengths • Each wavefunction is a standing wave that has zero amplitude at each end of the box – For n=2, the wavefunction has one complete cycle in the box – The number of nodes increases as n increases • The wavefunction is not an eigenfunction of the linear momentum operator. • However, each wavefunction is a superposition of functions with definite linear momenta – p = nh/2L or -nh/2L – particles can travel in opposite directions – equal probability for a particle traveling to the right or to the left Chapter 8 6 Particle in a Box – Properties of the Solutions • E1 = h2/8mL2 – E1 is the energy of the lowest energy state (n=1) – E1 is called the zero-point energy • En+1 – En = (2n +1)h2/8mL2 – The energy separation between energy levels (n+1) and n – The energy separation between adjacent levels increases as n increases – The energy separation decreases as length of container increases – The energy separation is very small when the container has macroscopic dimensions Chapter 8 7 Particle in a Box – The Probability Density • The probability density for a particle in a box is given by: • * = 2(x) = 2/L sin2(nx/L) • For a given wavefunction, n, we can by integration calculate the probability that the particle is found in a given region of the box • 2(x) becomes more uniform as n increases – The quantum mechanical result corresponds to the classical prediction at high quantum numbers ( the correspondence principle) Chapter 8 8 Orthogonality • Two wavefunctions are orthogonal if the integral of their product is zero • n and m are orthogonal if n*m d = 0 – Integrated over all space • Wavefunctions corresponding to different energies are orthogonal – For example n=1 and n=3 are orthogonal – Orthogonality is important in chemical bonding and spectroscopy Chapter 8 9 Harmonic Oscillator • A mass connected to a spring is a good example of a Harmonic oscillator (1degree of freedom). We define the displacement of the mass from the equilibrium position, Re, by the displacement coordinate, x • x = R - Re • Assume the force acting on the particle for small displacements is given by Hooke’s Law: F = -kx where k is the force constant • A negative sign is used because this is a restoring force. It acts in the opposite direction to the displacement • The force is related to the potential energy by: F = -dV/dx • Thus, dV/dx = kx Chapter 8 10 Harmonic Oscillator • The potential energy is given by: • V(x) = ∫ kx dx = ½kx2 • The Hamiltonian in the displacement coordinate is: H = px2 + V(x) • The Schrodinger equation can be written: • [(-ћ2/2m)(d2/dx2) + kx2/2]ψ = Eψ • The reduced mass, µ, of a molecule AB is given by • µ = mAmB /(mA + mB) Chapter 8 11 Energy Levels of a Harmonic Oscillator • Quantization of energy levels arises from the boundary conditions: • The oscillator will not be found with infinitely large compressions or extensions ( ψ = 0 at x = ±∞ ) • The permitted energy levels are: • Ev = (v + ½) hω v = 0, 1, 2, 3,….. • ω (the angular frequency) is given by: • ω = (k/m)½ or ω = (k/µ)½ (k is the force constant and µ the reduced mass) • ω = 2πυ υ is the vibrational frequency (periods per second) • Therefore, Ev = (v + ½) hυ v= 0, 1, 2, 3, …….. Chapter 8 12 Energy Levels of Harmonic Oscillator • The smallest permitted value of v is zero • The corresponding energy, E0 , is called the zero-point energy • E0 = ½hω = ½hυ • An oscillating particle is always fluctuating around its equilibrium position according to quantum mechanics. Classical mechanics would allow the particle to be perfectly still • The separation between adjacent energy levels is given by: • ΔE = Ev+1 – Ev = hυ • Same separation for all values of v Chapter 8 13 Wavefunctions for the Harmonic Oscillator • Form of wavefunctions (Solutions of Schrodinger equation): • ψ(x) = N (polynomial in x) (bell-shaped Gaussian Function) • N is a normalization constant • Gaussian function is of form exp(-x2) • The precise form of the wavefunctions are: • ψv(x) = Nv Hv(y) exp(-y2/2) • Here, y = x/α and α = (h2/mk)1/4 • Hv(y) is a Hermite polynomial • For ground state (v=0), H0(y) = 1 • Thus, ψ0(x) = N0 exp(-x2/2α2) • ψ0(x) and ψ02(x) are bell-shaped Gaussian functions Chapter 8 14 Wavefunctions and Probabilities for an Harmonic Oscillator • The wavefunction , ψv , has v nodes • Wavefunctions extend beyond the classical turning points (tunnelling – typical quantum mechanical results) • For v=0 (the ground state), the probability density is the highest at x=0 (Classical probability: max at the turning points for any energy) • For large v-values, the highest probability is near the classical turning points (The Correspondence Principle) Chapter 8 15 Rotational Motion - Angular Momentum • Particle of mass m rotating in a circular path of radius r • J = I· • J is the angular momentum of the particle – J can be represented by a vector perpendicular to the rotational plane • I is the moment of inertia – I = mr2 • is the angular velocity in radians per second Chapter 8 16 Rotation in Two Dimensions • Assume the particle is rotating in the xy-plane and that the potential energy, V, is zero • E = p2/2m – E is the total energy (kinetic energy only) – p is the linear momentum • Jz = ±pr – rotation in opposite directions give angular momenta of opposite sign • E = Jz2/2I – I = mr2 Chapter 8 17 Quantization of Rotational Motion • The wavefunction describing the particle must reproduce itself on successive circuits • Requirement: 2r = ml – ml is an integer – is the wavelength of the particle • Using the de Broglie relation ( = h/p ), it follows: • Jz = ћml (ml = 0, ±1, ±2, ±3, …….. ) – the angular momentum is quantized – positive values of ml correspond to rotation in clockwise direction – negative values of ml correspond to rotation in counter-clockwise direction Chapter 8 18 Rotation in Two Dimensions - Quantization of Energy • E = Jz2/2I = ml2ћ2/2I – ћ = h/2 – ml = 0, ±1, ±2, ±3, …… • Each ml-value gives a certain discrete energy – the energy is quantized Chapter 8 19 Rotation in Three Dimensions • Particle free to move on the surface of a sphere of radius r • Requirement: Wavefunction should match as a path is traced over the poles as well as around the equator of the sphere • A second cyclic boundary condition • A second quantum number is needed • Quantum numbers: l and ml • Possible values: l = 0, 1, 2, 3, ……… • ml = l, l-1, .., 0, …, -l+1, -l • For a given l-value, there are (2l+1) permitted values of ml Chapter 8 20 Wavefunctions and Energies for Rotational Motion • The normalized wavefunctions, Yl,m(,), are called spherical harmonics • E = l(l+1) ћ2/2I – l = 0, 1, 2, 3, …… – ћ = h/2 • The energy, E, is quantized • E is independent on ml – there are (2l+1) wavefunctions foreach l-value. The energy level corresponding to a particular l-value is (2l+1) degenerate Chapter 8 21 Angular Momentum • J =[l(l+1)]½ћ • J is the magnitude of the angular momentum • z-component of angular momentum = mlh – ml = l, l-1,…,0, ….., -l – orientation of a rotating body is quantized - space quantization • From the l-value, one can calculate the orbital angular momentum of an electron orbiting around the nucleus • One can also calculate the z-component of the orbital angular momentum ( mlћ ) – ml is called the orbital magnetic quantum number Chapter 8 22 Electron Spin • Electrons behave as if each is spinning around its own axis • The electron spin quantum number, s = ½ • The magnitude of the spin angular momentum = [s(s+1)]½ ћ – ћ = h/2 • The electron can have two spin directions, ms = +½ or -½ – ms is called the spin magnetic quantum number • The z-component of the spin angular momentum is msћ Chapter 8 23

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1 milliliter, Avogadro's number, significant figures, translational motion, electrical wiring, Classical Mechanics, Assignment 1, dimensional analysis, Atomic Mass Unit, AP Physics C

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posted: | 3/24/2011 |

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