3502_04_sept16_LG

3502_04_sept16_LG - Chem 3502/5502 Physical Chemistry II...

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Chem 3502/5502 Physical Chemistry II (Quantum Mechanics) 3 Credits Fall Semester 2009 Laura Gagliardi Lecture 4, September 16, 2009 Solved Homework If an electron has a de Broglie wavelength of 1 Å (0.1 nm), then we can compute its momentum as p = h ! = 6.626 " 10 # 34 Js 1.0 " 10 # 10 m = 6.626 " 10 # 24 kgms -1 The kinetic energy of an electron having this momentum is T = p 2 2 m = 6.626 ! 10 " 24 kg ms ( ) 2 2 ! 9.11 ! 10 " 31 kg = 2.41 ! 10 " 17 J The energy of an electron subjected to one volt of accelerating potential is 1 eV. The conversion from eV to J is 1 eV = 1.602 x 10 19 J. The kinetic energy of the electron is thus 150.4 eV. So, an accelerating potential of roughly 150 V is required. The Wave Function for a Material System Erwin Schrödinger, in 1926, proposed a way to meld the many quantum observations up to that point with a wavelike description of matter. In particular, he started from the classical wave equation (shown here for a single dimension) ! 2 " x , t ( ) ! x 2 = 1 c 2 ! 2 " x , t ( ) ! t 2 (4-1) where Ψ is a wave function which has an amplitude for any specification of x (position) and t (time), and c is the velocity of the wave (wavelength times frequency). A general solution to this equation is ! x , t ( ) = Ce 2 " i x # $% t & ( ) * + (4-2) where C is an arbitrary multiplicative constant, i is the square root of negative one (a base for the complex numbers), λ is the wavelength and ν is the frequency. To verify, note that ! 2 " x , t ( ) ! x 2 = # 4 C $ 2 % 2 e 2 $ i x % #& t ( ) * + , (4-3)
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4-2 and ! 2 " x , t ( ) ! t 2 = # 4 C $ 2 % 2 e 2 $ i x & #% t ( ) * + , (4-4) and c = λν . Schrödinger decided to use the de Broglie wavelength for λ and to relate the frequency to the Planck energy. That is, he looked for a differential equation having the solution ! x , t ( ) = Ce i xp " Et h # $ % & ( (4-5) where p is the momentum and E is the energy. For notational simplicity, from now on we will take the arbitrary constant C as one. If we consider differentiating eq. 4-5 once with respect to time we have !" x , t ( ) ! t = # iE h e i xp # h $ % & ( ) (4-6) or ! h i "# x , t ( ) " t = E # x , t ( ) (4-7) The total energy E at a given time may be expressed as a sum of kinetic and potential energy, both of which depend only on x and not on t . Thus, we have ! h i x , t ( ) " t = p 2 2 m + V x ( ) $ % & ( ) # x , t ( ) (4-8) where p is the momentum, m is the mass, and V is the potential energy, which may be thought of as an outside influence on the system being described by the wave function Ψ . Note that the idea here is to have Ψ be a function that contains information about the system contained within it. That is, we would like things like momentum and energy to be themselves determinable from the wave function. Let us consider how we might determine the momentum from our wave function eq. 4-5. If we differentiate once with respect to x we have x , t ( ) ! x = ip h e i xp # h $ % & ( ) (4-9)
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4-3 We may write this in what is known as "operator" formalism as ! ! x " x , t ( ) = ip h " x , t ( ) (4-10) or h i ! ! x = p (4-11) which is to say that the operator that can be applied to the wave function in order to determine the momentum is to differentiate once with respect to x and then multiply by h-bar over i . In this case, it is straightforward to show that the square of the momentum (needed for the kinetic energy) can be derived from operating twice
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This note was uploaded on 12/14/2010 for the course CHEM 3502 taught by Professor Staff during the Spring '08 term at Minnesota.

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3502_04_sept16_LG - Chem 3502/5502 Physical Chemistry II...

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