# Chap 35 SM - Chapter 35 Applications of the Schrdinger...

This preview shows pages 1–4. Sign up to view the full content.

57 Chapter 35 Applications of the Schrödinger Equation Conceptual Problems 1 Sketch ( a ) the wave function and ( b ) the probability density function for the n = 5 state of the finite square-well potential. 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 ) 0 ψ 5 x L ( b ) 0 L 5 2 x 2 Sketch ( a ) the wave function and ( b ) the probability density function for the n = 4 state of the finite square-well potential. 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 ) x 4 0 L ( b ) x L 0 4 2

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Chapter 35 58 The Schrödinger Equation 3 •• Show that if ψ 1 ( x ) and 2 ( x ) are each solutions to the time-independent Schrödinger equation (Equation 35-4), then 3 ( x ) = 1 ( x ) + 2 ( x ) is also a solution. This result, known as the superposition principle, applies to the solutions of all linear equations. Picture the Problem We can show that 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 = 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 = + = and ( ) x E x x U dx x d m 2 2 2 2 2 2 2 = + = 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 + = + + + = (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 = + = which shows that ) ( ) ( ) ( 2 1 3 x x x + = satisfies Equation 35-4. The Harmonic Oscillator 4 •• The harmonic oscillator problem may be used to describe the vibrations of molecules. For example, the hydrogen molecule H 2 is found to have equally spaced vibrational energy levels separated by 8.7 × 10 –20 J. What value of the force constant of the spring would be needed to get this energy spacing,
Applications of the Schrödinger Equation 59 assuming that half the molecule can be modeled as a hydrogen atom attached to one end of a spring that has its other end fixed? Hint: The spacing for the energy levels of this half-molecule would be half of the spacing for the energy levels of the complete molecule. In addition, the force constant of a spring is inversely proportional to its relaxed length, so if half of the spring has force constant k , the entire spring has a force constant that is equal to 1 2 k . 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 force constant k.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 02/28/2012 for the course PHYSICS 122 taught by Professor Goussiou during the Spring '12 term at University of Washington.

### Page1 / 42

Chap 35 SM - Chapter 35 Applications of the Schrdinger...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online