{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW 03(1) - EEE 352Spring 2013 Homework 3Due January 28 2.12...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
EEE 352—Spring 2013 Homework 3—Due January 28 2.12 When the uncertainty principle is considered, it Is not possible to locate a photon in space more precisely than about one wavelength. Consider a photon with wavelength λ = 1 μ m. What is the uncertainty in the photon’s momentum? From the problem definition, we take Δ x = 10 6 m and from the footnote on p. 30, we have Δ p = 2 Δ x = 5.27 × 10 29 kg m / s . 2.18 Consider the wave function Ψ ( x ) = A cos n π x ( ) e i ω t , 1 2 x 1 2 . Determine A such that the wave function is properly normalized . The normalization is given such that 1 = Ψ ( x ) 2 dx 1/2 1/2 = A 2 cos n π x ( ) e i ω t 2 dx 1/2 1/2 = A 2 cos 2 n π x ( ) dx 1/2 1/2 = A 2 2 1 + cos 2 n π x ( ) $ % & ' dx 1/2 1/2 = A 2 2 x + 1 2 n π sin 2 n π x ( ) $ % ( & ' ) 1/2 1/2 = A 2 2 A = 2 . 2.20 An electron is given by a wave function given by ψ ( x ) = 2 a cos π x a ! " # $ % & , a 2 x a 2 . The wave function is zero elsewhere. Calculate the probability of finding the electron between (a) 0 < x < a/4, (b) a/4 < x < a/2, (c) –a/2 < x < a/2. The expectation value of x is given by
Image of page 1

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

View Full Document Right Arrow Icon
x = 2 a cos π x a ! " # $ % & x cos π x a ! " # $ % & dx x 1 x 2 = 1 a 1 + cos 2 π x a ! " # $ % & ( ) * + , - xdx x 1 x 2 = 1 a x x 1 x 2 x a a 2 π sin 2 π x a ! " # $ % & x 1 x 2 + 1 2 π sin 2 π x a ! " # $ % & dx x 1 x 2 = x 2 x 1 a 1 2 π x 2 sin 2 π x 2 a ! " # $ % & x 1 sin 2 π x 1 a ! " # $ % & ( ) * + , - + 1 2 π ( ) a 2 π cos 2 π x a ! " # $ % & x 1 x 2 = x 2 x 1 a 1 2 π x 2 sin 2 π x 2 a ! " # $ % & x 1 sin 2 π x 1 a ! " # $ % & ( ) * + , - + a 2 π ( ) 2 cos 2 π x 2 a
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern