{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

quiz2solution

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

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

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:

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.

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

• 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.

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

• 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.

Dana University of Pennsylvania ‘17, Course Hero Intern

• 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.

Jill Tulane University ‘16, Course Hero Intern