{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

quiz2solution - Square-barrier tunneling T" 16 E U o E...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
EE 439 Name_______________________________________ Quiz 2 – Sept. 21, 2009 Put your final answers on this sheet and attach any additional sheets behind. You must include your work to get full credit. An proton (mass = 1.67x10 –27 kg) with energy of 100 eV is incident on a square barrier with energy of 110 eV. Calculate the thickness that the barrier must have in order for the tunneling probability to be 1%? Inverting the probability equation for barrier thickness, L, L = " 1 2 # ln TU o 2 16 E U o " E ( ) $ % & ( ) . " = 2 m U o # E ( ) h 2 = 2 1.67 $ 10 # 27 kg ( ) 110 eV # 100 eV ( ) 1.6 $ 10 # 19 J / eV ( ) 1.055 $ 10 # 34 J % s ( ) 2 = 6.93 $ 10 11 m # 1 = 693 nm # 1 Then, L = " 1 2 693 nm " 1 ( ) ln 0.01 ( ) 110 eV ( ) 2 16 100 eV ( ) 110 eV " 100 eV ( ) # $ % % & ( ( = 0.0035 nm Potentially useful equations and data: Square-barrier tunneling:
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
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Square-barrier tunneling: T " 16 E U o # E ( ) U o 2 exp # 2 $ L ( ) ; = 2 m U o # E ( ) h 2 Energy levels in an infinitely deep quantum well: E n = n 2 " 2 h 2 2 mL 2 p = h E = h # "# = c h = 6.63x10 –34 J·s = 4.14x10 –15 eV·s; h = 1.055x10 –34 J·s = 6.595x10 –16 eV·s (Planck’s constant) c = 3.0x10 8 m/s (speed of light in vacuum) k = 1.38x10 –23 J/K = 8.617x10 –5 eV/K (Boltzmann’s constant, not wave number) m = 9.11x10 –31 kg (mass of an electron) q = 1.6x10 –19 C (charge of an electron) ε o = 8.85x10 –12 F/m (free-space permittivity) 1 eV = 1.6x10 –19 J...
View Full 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