Class Notes (838,472)
United States (325,435)
Physics (56)
PH 122 (46)
Lecture

Heisenberg uncertainty principle

3 Pages
32 Views
Unlock Document

Department
Physics
Course
PH 122
Professor
Brian Jones
Semester
Spring

Description
2 April Light of wavelength 400 nm illuminates a potassium electrode (work function 2.3 eV) What is the photon energy? 1240 E photon λ = 3.1 eV What is the energy of the emitted electron? 3.1 eV – 2.3 eV = 0.8 eV What is the stopping potential? -0.8 V Ocean water is most transparent at wavelengths of 470 nm, so bioluminescent creatures emit light at approximately this wavelength. Firefly squid use ATP to provide the energy for this reaction. Metabolizing one molecule of ATP releases 0.32 eV. How many molecules of ATP must be metabolized to produce one photon of blue light at 470 nm? 1240 1240 E photon = = 2.6 eV λ 470nm 2.6eV 0.32eV = 8.15 9 molecules of ATP Electrons are accelerated from rest through an 8000 V potential difference. By what factor would their de Broglie wavelength increase if they were instead accelerated through a 2000 V potential? K = Δ Ue K = ½ mv 2 h h λ = p = mv ΔU ↓ 4 K ↓ 4 v ↓ 2
More Less

Related notes for PH 122

Log In


OR

Join OneClass

Access over 10 million pages of study
documents for 1.3 million courses.

Sign up

Join to view


OR

By registering, I agree to the Terms and Privacy Policies
Already have an account?
Just a few more details

So we can recommend you notes for your school.

Reset Password

Please enter below the email address you registered with and we will send you a link to reset your password.

Add your courses

Get notes from the top students in your class.


Submit