5.111 Lecture Summary #4 Wednesday, September 10, 2014
Reading for today: Section 1.5 and Section 1.6. (Same sections in 5th and 4th editions)
Read for Lecture #5: Section 1.3 – Atomic Spectra, Section 1.7 up to equation 9b –
Wavefunctions and Energy Levels, Section 1.8 – The Principle Quantum Number.
(Same sections in 5th and 4th editions)
Pre-lecture questions are a learning tool for you to test your knowledge of the
material. They will not be graded on correctness, but rather on competition.
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Topics: I. Light as a particle continued
II. Matter as a wave
III. The Schrödinger equation
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I. LIGHT AS A PARTICLE CONTINUED
A) More on the Photoelectric Effect
e- ejected e- ejected e- ejected
Three photons, each with an energy equal to φ/2 eject an electron!
Terminology tips to help solve problems involving photons and electrons:
• photons: also called light, electromagnetic radiation, etc.
o may be described by , , or
• electrons: also called photoelectrons.
o may be described by , , or λ (see part II of today’s notes)
• eV is a unit of energy = 1.6022 x 10-19 J.
1
, NOW FOR AN IN-CLASS DEMO OF THE PHOTOELECTRIC EFFECT:
Metal surface: Zn, φ =
Incident light sources:
• UV lamp with a λ centered at 254 nm
• Red laser pointer (λ = 700 nm)
First, let’s solve the following problems to determine if there is sufficient energy in a single
photon of UV or of red light to eject an electron from the surface of the Zn plate. For
calibration, we’ll also calculate the # of photons in a beam of light.
Consider our two light sources: a UV lamp (λ = 254 nm) and a red laser pointer (λ = 700.
nm).
1) What is the energy per photon emitted by the UV lamp?
2) What is the energy per photon emitted by the red laser pointer?
3) What is the total number of photons emitted by the laser pointer in 60 seconds if the
intensity (I) = 1.00 mW?
1) What is the energy per photon emitted by the UV lamp? λ = 254 nm
E= ν= E=
E= E=
The UV lamp have enough energy per photon to eject electrons from the
surface of a zinc plate (φ of Zn = 6.9 x 10-19 J).
2) What is the energy per photon emitted by the red laser? λ = 700. nm
E=
E= (6.626 x10-34 Js)(2.998 x 108 m/s) E=
The red laser have enough energy per photon to eject electrons from the
surface of a zinc plate (φ of Zn = 6.9 x 10-19 J).
2
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