physicsa.com
 class code: SPS22 teacher: Mr. Elert classroom: A314 office: A214 test day: Friday phone: (718) 724–8500 ext. 2141 email: elert@midwoodscience.org

## Physics A: Problem Set 22: The Particle Nature of Light

### recommended reading

 Barron's Let's Review: 13.1–13.4 physics.info: Photoelectric Effect, X-rays Wikipedia: Photoelectric effect, Photon, X-ray HyperPhysics: Photoelectric Effect, Wave-Particle Duality Khan Academy: Photoelectric effect

### in class

1. Complete the following table for photons of visible light.
λmax λmed λmin
wavelength (nm) 700 nm 550 nm 400 nm
frequency (Hz)
energy per photon (J)
energy per photon (eV)
color
λmax λmed λmin
wavelength (m) 700 nm 550 nm 400 nm
frequency (Hz) 4.28 × 1014 Hz 5.45 × 1014 Hz 7.49 × 1014 Hz
energy per photon (J) 2.84 × 10−19 J 3.61 × 10−19 J 4.97 × 10−19 J
energy per photon (eV) 1.77 eV 2.26 eV 3.10 eV
color red green violet

Frequency is computed using the wave speed equation and the speed of light in a vacuum (c = 3.00 × 108 m/s).

 f = c ⇐ c = fλ λ

Energy per photon in joules is computed using Planck's relationship and Planck constant (h = 6.63 × 10−34 Js).

E = hf

To convert from joules to electronvolts, divide the energy in joules by the elementary charge in coulombs (q = 1.60 × 10−34 C).

 V = E q

The visible spectrum starts with red and ends with violet when the colors are listed in order of increasing frequency (red, orange, yellow, green, blue, violet). Since frequency and wavelength are inversely proportional, the lowest frequency light (red) goes with the longest wavelength and the highest frequency light (violet) goes with the shortest wavelength and green is in the middle.

2. Complete the following table for different types of electromagnetic radiation.
E1 E2 E3
energy per photon (eV) 100 eV 103 eV 106 eV
energy per photon (J)
frequency (Hz)
wavelength (m)
type of radiation
E1 E2 E3
energy per photon (eV) 100 eV 103 eV 106 eV
energy per photon (J) 1.60 × 10−19 J 1.60 × 10−16 J 1.60 × 10−13 J
frequency (Hz) 2.42 × 1014 Hz 2.42 × 1017 Hz 2.42 × 1020 Hz
wavelength (m) 1.24 × 10−6 m 1.24 × 10−9 m 1.24 × 10−12 m
type of radiation infrared x‑ray
(ultraviolet)
gamma ray
(x‑ray)

To convert from electronvolts to joules, multiply the voltage in volts by the elementary charge in coulombs (q = 1.60 × 10−34 C).

 E = qV ⇐ V = E q

Frequency is computed using Planck's equation and Planck's constant (h = 6.63 × 10−34 Js).

 f = E ⇐ E = hf h

Wavelength is computed using the wave speed equation and the speed of light in a vacuum (c = 3.00 × 108 m/s).

 λ = c ⇐ c = fλ f

The type of radiation can be found in the spectrum chart of your choice.

3. The Balmer series is a set of wavelengths emitted by excited hydrogen atoms. Complete the following table for the four visible wavelengths in the Balmer series identified by the Greek letters α (alpha), β (beta), γ (gamma), and δ (delta).
α β γ δ
wavelength (nm) 656 nm 486 nm 434 nm 410 nm
frequency (Hz)
energy per photon (J)
energy per photon (eV)
color
α β γ δ
wavelength (nm) 656 nm 486 nm 434 nm 410 nm
frequency (Hz) 4.57 × 1014 Hz 6.17 × 1014 Hz 6.91 × 1014 Hz 7.32 × 1014 Hz
energy per photon (J) 3.03 × 10−19 J 4.09 × 10−19 J 4.58 × 10−19 J 4.85 × 10−19 J
energy per photon (eV) 1.90 eV 2.56 eV 2.86 eV 3.03 eV
color red blue violet violet

Frequency is computed using the wave speed equation and the speed of light in a vacuum (c = 3.00 × 108 m/s).

 f = c ⇐ c = fλ λ

Energy per photon in joules is computed using Planck's relationship and Planck constant (h = 6.63 × 10−34 Js).

E = hf

To convert from joules to electronvolts, divide the energy in joules by the elementary charge in coulombs (q = 1.60 × 10−34 C).

 V = E q

Color is found using your friendly neighborhood reference tables. If you don't have a reference table, I hear there's this thing called Google that can help you find one.

### at home

1. Three related questions.
1. Why are ultraviolet, x-rays, and gamma rays always regarded as harmful while infrared, microwaves, and radio waves are generally regarded as benign (by sensible people, anyway).
2. In what sense can visible light be thought of as "just right" as an energy source for life (and other forms of chemistry) on earth?
3. Under what circumstances could exposure to the generally benign electromagnetic waves listed in part a. be considered harmful?

Three related answers.

1. Radiation on the high end of the electromagnetic spectrum (ultraviolet, x-rays, gamma rays) are always regarded as harmful since their energy per photon is high enough to ionize electrons, dissociate molecules, or even transform nuclei — events that mess with the basic chemistry of life as we know it. Radiation on the low end of the electromagnetic spectrum (infrared, microwaves, radio waves) are generally regarded as benign since their energy per photon is too low to do these things.
2. Visible light can be thought of as "just right" since photons in this part of the electromagnetic spectrum have just enough energy to excite electrons, but not so much that they would kick them out of the atom. Photosynthesis, the fundamental means of extracting energy for life on earth, relies on excited electrons as a means to drive chemical reactions. Ultraviolet and higher frequencies can only dissociate molecules. Infrared and lower frequencies can only vibrate molecules.
3. Exposure to non-ionizing radiation at high intensity can cause burns. Let's use microwaves as an example. Microwaves are low frequency electromagnetic waves that will not ionize atoms or dissociate molecules, which is why they are generally regarded as safe. When microwaves are absorbed by living tissues, heat is transferred. At low intensities they can be used to soothe a sore back in much the same way as a hot water bottle or electric heating pad. At high intensities, they can be used to boil water for tea or coffee. Were someone foolish enough to dismantle the safety mechanism on a microwave oven, turn it on, and then place a hand inside, they would find themselves cooking their own flesh. That is an example of the harmful use of an otherwise safe device.
2. Pigments are used to give color to paint (on a house or on an artist's canvas), fibers (in clothing, carpeting, or tapestries), and glaze (on ceramic pottery or mosaic tiles).
1. Why does red pigment appear red? Why does blue pigment appear blue?
2. How does the energy of a photon of red light compare to the energy of a photon of blue light?
3. Which pigment is more likely to fade first when exposed to sunlight, red or blue? Justify your answer.
1. Red pigment appears red because it reflects red photons and absorbs blue photons. Similarly blue pigment appears blue because it reflects blue photons and absorbs red photons.
2. Red photons have a lower energy than blue photons since energy is directly proportional to frequency as shown in Planck's equation E = hf.
3. Red pigment is more likely absorb higher energy photons than blue pigment. Higher energy photons are more likely to cause chemical changes when absorbed than lower energy photons. The color of a pigment is a chemical property. Thus, red pigment tends to fade faster than blue pigment.