Applications of the Photon Model - Solutions

Hi everyone,

I want to let you know about a typo in #7 on the sheet. The shift in wavelength should be given by 2h/m_ec.

1997B Part II.
a)
b) The energy must be -1.9 eV, based on the energy emitted by the 600 nm photon.
c) 1240 nm - outside of the visible spectrum


200...Something, #7:
a) There are actually two (correct) ways to draw the diagram:

b) 1240 nm, again, outside of the visible spectrum.


7.
a) 9.9 x 10^-15 J
b) 3.3 x 10^-23 kg*m/s
c) The photon wavelength increases because some of the energy of the incident photon was lost to KE of the electron.
d) 6.0 x 10^-23 kg*m/s

1993B6
a)1.69 x 10¹⁹ Hz
b)3.73 x 10^-23 kg*m/s
c)3.31 x 10^-16 J
d)2.46 x 10^-23 kg*m/s

Quantum Theory, Photoelectric Effect and X-Rays

Photoelectric Effect
http://www.lon-capa.org/~mmp/kap28/PhotoEffect/photo.htm

More visual Photoelectric effect:
http://phet.colorado.edu/simulations/photoelectric/photoelectric.jnlp

Solutions:
2000B5
a)
i) 4.5 eV or 7.2 x 10^-19 J
ii) 1.26 x 10⁶ m/s
b) 183 nm
c) 5.56 x 10¹⁴ Hz

1980B3
a) The graph should be straight forward, but you SHOULD know that all four points cannot be on the same line!
b) 0.75 x 10¹⁴ Hz
c) 3.1 eV
d) Energy of electrons is 5 eV according to data, so the stopping potential will be 5 volts.
e) Charges move in circular paths when they are in a magnetic field perpendicular to their velocity. This is a possible reason for this behavior.

Page 855:
16. 4.14 x 10¹⁴ Hz

17. 401 nm

19. The shortest visible wavelength photons have energies of 3.1 eV (400 nm wavelength). Copper and Iron have work functions that are greater than this, so they will not eject photo electrons in visible light.
20.

a) 3.49 x 10^-19 J or 2.179 eV.
b) 0.930 V

Page 885:
P32. 4.14 x 10^-11 m (Don't worry about the longest wavelength.)
P33. 41,000 V

Diffraction, Single and Double Slit Solutions

From Handout:
1.
a) 2.5 mm
b) wavelength = 385 nm, frequency = 6 x 10¹⁴ Hz
c) Since the wavelength decreases, and the spacing between maxima is proportional to wavelength, the space between the maxima will decrease.

1991B6.
a) 3.9 x 10^-5 m
b) 9.6 x 10^-3 m

Problems:
7. 8.5 x 10^-5m
10. 613 nm
20. 1.57 mm 1.591 meters
21. Notice that this says the distance between maxima, not minima. The path difference must be half a wavelength more than that required for the minimum. The total angle is 19.5 degrees across the central maximum, so the angle between the central maximum and the first minimum is 9.75 degrees. sin(9.75) = 1.5*lambda/W, so W = 5606 nm.

Thin Film Interference Solutions

From handout:

4. thickness = 1.05 x 10^-7 m

5. thickness = 96.1 nanometers

1984B5
a) 5 x 10¹⁴ Hz
b) 4.8 x 10^-7 m
c) 1.2 x 10^-7 m
d) 2.4 x 10^-7 m

38. wavelength = 643 nm, which is colored red.
40. 169 nm
43. 27*lambda/n = 2t, so t = 9045 nanometers, or 9.045 x 10^-6 m
45. 2t = n*(640 nm/1.36), and 2t = (n + 1/2)*512 nm/1.36. Solving the system, t = 471 nanometers.

Diffraction, Single and Double Slits

Huygen's Principle
http://projects.cbe.ab.ca/sss/science/physics/map_north/applets/advancedripple/advancedripple.html

Single Slit Simulation:
http://www.physics.uoguelph.ca/applets/Intro_physics/kisalev/java/slitdiffr/index.html

Double Slit:
http://www.walter-fendt.de/ph14e/doubleslit.htm

Ripple Tank
http://www.falstad.com/ripple/

Refraction Simulations

Total Internal Reflection Simulation:
http://www.physics.uoguelph.ca/applets/Intro_physics/kisalev/java/totintrefl/index.html

Prism/Dispersion
http://www.physics.uoguelph.ca/applets/Intro_physics/kisalev/java/dispprizm/index.html

Glass Slab Refraction
http://www.physics.uoguelph.ca/applets/Intro_physics/kisalev/java/dispslab/index.html

Lenses - Solutions

Multiple Choice problems from the handout:
1. B
2. E
3. C

Questions:
20. If the object distance is very large, then by the thin lens equation, 1/d0 approaches zero. This means the image distance is equal to the focal length. This is where the film must be placed to form an image.

Problems:
48b. 390 mm

50. Since the image is on the other side of the lens, it must be a real image. This means that the image distance is positive. Using the thin lens equation, the focal length must be 39.8 cm.

52. 72 cm on the same side of the lens as the stamp (virtual image, negative image distance). Magnification = -4

60.
a) 15.2 cm, real image (opposite side of the lens as object), height is -2.78 mm
b) -12.1 cm, virtual image (negative image distance), height is +2.22 mm

83. Since a diverging lens has a negative focal length, by the thin lens equation, 1/di can never be positive. If di is negative, the image is on the same side as the object, so it will be virtual. The only way for di to be positive is if d0 for the object is negative. This can occur, and we will discuss HOW it can occur in class Monday!



AP Problems:
1989B6.
a) real
b) di = 30 cm
c) M = -0.5
d) See the image to the left.
e) focal length decreases - by Snell's law, the rays will be bent more as they enter and exit the lens.













1986B5
a) 7.4%
b) 1.250 x 10⁶ W
c) We'll get to this problem in a couple weeks....
d)