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Chapter 36

# Chapter 36 - CHAPTER 36 Applications of the Schrdinger...

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CHAPTER 36 Applications of the Schrödinger Equation 1* True or false: Boundary conditions on the wave function lead to energy quantization. True 2 Sketch ( a ) the wave function and ( b ) the probability distribution for the n = 4 state for the finite square- well potential. ( a ) The wave function is shown below ( b ) The probability density is shown below 3 Sketch ( a ) the wave function and ( b ) the probability distribution for the n = 5 state for the finite square- well potential. ( a ) The wave function is shown below ( b ) The probability density is shown below

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Chapter 36 Applications of the Schrödinger Equation 4 ∙∙ Show that the expectation value < x > = x Ψ 2 dx is zero for both the ground and the first excited states of the harmonic oscillator. The integral 0 2 = dx x ψ because the integrand is an odd function of x for the ground state as well as any excited state of the harmonic oscillator. 5* ∙∙ Use the procedure of Example 36-1 to verify that the energy of the first excited state of the harmonic oscillator is E 1 = 2 3 h _ ω 0 . ( Note : Rather than solve for a again, use the result a = m 0 /2 h _ obtained in Example 36-1.) The wave function is e x A = ax 2 1 (see Equ. 36-25). Then e A ax e A = dx d ax ax 1 2 2 1 2 2 and e A ax) x a = e A x a + e axA e axA = dx d ax ax ax ax 2 2 2 2 1 3 2 1 3 2 1 1 2 2 6 (4 4 4 2 . We now substitute this into the Schrödinger equation. The exponentials and the constant A 1 cancel, so ( h _ 2 /2 m )(4 a 2 x 3 – 6 ax ) + 1/2 m 0 2 x 3 = E 1 x . With a = 1/2 m 0 / h _ , the terms in x 3 cancel, and solving for the energy E 1 we find E 1 = 6 h _ 2 a /2 m = 3 h _ 0 /2 = 3 E 0 . 6 ∙∙∙ Show that the normalization constant A 0 of Equation 36-23 is A 0 = (2 m 0 / h ) 1/4 . Let A 0 = (2 m 0 / h ) 1/4 = ( m 0 / π h _ ) 1/4 . Then dx x 2 0 ) ( = ( m 0 / h _ ) 1/2 exp[(– m 0 / h _ ) x 2 ] dx = = 1 2 2 / 1 ds e s . 7 ∙∙∙ Find the normalization constant A 1 for the wave function of the first excited state of the harmonic oscillator, Equation 36-25. We require that = 1 2 2 2 2 1 dx e x A ax where a = m 0 /2 h _ . Let y 2 = 2 ax 2 ; then the integral becomes = 2 / 3 2 2 / 3 ) 2 ( 2 ) 2 ( 2 a dy e y a y . Consequently, a = A 3 2 1 32 and h m = A 1/4 3 2 3 0 3 1 32 . 8 ∙∙∙ Find the expectation value < x 2 > = x 2 2 dx for the ground state of the harmonic oscillator. Use it to show that the average potential energy equals half the total energy. = dx e x A dx x ax 2 2 2 2 0 2 2 . The integral has already been evaluated in Problem 36-7. Using that result and
Chapter 36 Applications of the Schrödinger Equation A 0 given in Problem 36-6, x 2 = h _ /2 m ω 0 . The average potential energy of the oscillator is 1/2 m 0 2 x 2 = h _ 0 /4 = E 0 /2.

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Chapter 36 - CHAPTER 36 Applications of the Schrdinger...

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