ism_chapter_35

# Ism_chapter_35 - Chapter 35 Applications of the Schrdinger Equation Conceptual Problems 1 True 2 Determine the Concept Looking at the graphs in the

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1137 Chapter 35 Applications of the Schrödinger Equation Conceptual Problems 1 True 2 Determine the Concept Looking at the graphs in the text for the n = 1, 2, and 3 states, we note that the n = 4 state graph of the wave function must have four extrema in the region 0 < x < L and decay in toward zero in the regions x < 0 and x > L . ( a ) ( b ) 3 Determine the Concept Looking at the graphs in the text for the n = 1, 2, and 3 states, we note that the n = 5 state graph of the wave function must have five extrema in the region 0 < x < L and decay in toward zero in the regions x < 0 and x > L . ( a ) ( b )

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Chapter 35 1138 Estimation and Approximation *4 Picture the Problem Assume a mass of 150 g for the baseball, 30 cm for the width of the locker, and 1 cm/s for the speed of the ball, and equate the kinetic energy of the ball and the quantum-mechanical energy and solve for the quantum number n . The allowed energy states of a particle of mass m in a 1-dimensional infinite potential well of width L are given by: = 2 2 2 8 mL h n E n The kinetic energy of the ball is: 2 2 1 mv K = For E n = K : 2 2 2 2 2 1 8 mv mL h n = Solve for the quantum number n : h mvL n 2 = Substitute numerical values and evaluate n : ( )( )( ) 30 30 34 10 10 36 . 1 s J 10 63 . 6 m 3 . 0 m/s 01 . 0 kg 15 . 0 2 × = × = n The Schrödinger Equation 5 •• Picture the Problem We can show that ψ 3 ( x ) is a solution to the time-independent Schrödinger equation by differentiating it twice and substituting in Equation 35-4. Equation 35-4 is: ( ) () () () x E x x U dx x d m = + 2 2 2 2 h Because 1 ( x ) and 2 ( x ) are solutions of Equation 35-4: ( ) x E x x U dx x d m 1 1 2 1 2 2 2 = + h and ( ) x E x x U dx x d m 2 2 2 2 2 2 2 = + h
Applications of the Schrödinger Equation 1139 Add these equations to obtain: () [] ) ( ) ( ) ( ) ( ) ( ) ( 2 2 1 2 1 2 2 2 2 1 2 2 x x E x x x U dx x d dx x d m ψ + = + + + h (1) Differentiate ) ( ) ( ) ( 2 1 3 x x x + = twice with respect to x to obtain: dx x d dx x d dx x d ) ( ) ( ) ( 2 1 3 + = and 2 2 2 2 1 2 2 3 2 ) ( ) ( ) ( dx x d dx x d dx x d + = Substitute in equation (1) to obtain: ) ( ) ( ) ( 2 3 3 2 3 2 2 x E x x U dx x d m = + h which shows that ) ( ) ( ) ( 2 1 3 x x x + = satisfies Equation 35-4. The Harmonic Oscillator 6 •• Picture the Problem We can relate the spring constant to the mass of the hydrogen atom and its angular frequency and then use the relationship between the allowed energy levels and the angular frequency ω to derive an expression for the spring constant k. The spring constant k is related to the mass m of the hydrogen molecule and its angular frequency : 2 m k = (1) Relate the energy spacing E to the angular frequency : π h = = = 2 h hf E Solve for : h E = Substitute for in equation (1) to obtain: 2 ⎛ ∆ = h E m k Substitute numerical values and evaluate k :

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Chapter 35 1140 kN/m 14 . 1 s J 10 05 . 1 J 10 7 . 8 u kg 10 1.66 u 1 2 34 20 27 = × × × × = k Remarks: Our result is very similar to the stiffness constant of typical macroscopic springs.
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## This homework help was uploaded on 02/26/2008 for the course PHYSICS 11 taught by Professor Licini during the Spring '07 term at Lehigh University .

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Ism_chapter_35 - Chapter 35 Applications of the Schrdinger Equation Conceptual Problems 1 True 2 Determine the Concept Looking at the graphs in the

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