Waves/Interference Review Solutions:
1 D
2 A
3 C
4 C
5 B
6 A
7 D
8 E
9 C
10 E
11 C
12 D
13 A
14. B (the radical symbol should end before the f)
15. B
16. A
17. B
18. A
19. D
Preview for tomorrow...let's shed some light on MC optics questions....
Monday, April 20, 2009
Tuesday, April 14, 2009
Working Hard, or hardly working?
Hi everyone,
I want to thank those of you that have contacted me thus far. Those of you that have not...it's cranky time....(Azeezat, Maria, Danielle, Brandon, Sam, Marlon...did you lose my number?)
I've been working hard in Cleveland to get wedding details figured out - running around the city, picking things up, dropping them off, and of course, deciding on cake...here's the evidence of my "hard work" after dinner tonight:
Actually, I'm feeling a bit sick. I never thought I could have too much cake.
Thankfully, you can NEVER have enough physics.
Take care everyone,
EMW
I want to thank those of you that have contacted me thus far. Those of you that have not...it's cranky time....(
I've been working hard in Cleveland to get wedding details figured out - running around the city, picking things up, dropping them off, and of course, deciding on cake...here's the evidence of my "hard work" after dinner tonight:
Actually, I'm feeling a bit sick. I never thought I could have too much cake.
Thankfully, you can NEVER have enough physics.
Take care everyone,
EMW
Sunday, April 12, 2009
So You Think You Can Physics?
Tuesday, April 7, 2009
Is this really the end? I was just getting into it....
Problem 1.
a + b)
Don't worry about the fact that the signs are different from what we discussed in class today - your answers to the other parts are still consistent with the solutions below, and will be correct.
c) Highest energy is the 6 eV transition.
d) lowest is the 1 eV transition.
e) This is the reverse of c and d - longest wavelength will be the 1 eV transition since E = hc/lambda
f) shortest wavelength is with the 6 eV transition, which has the highest energy.
g) The difference in energy between these two energy levels is 5 eV. This translates into a photon of wavelength 248 nanometers.
h) Three are possible.
i) The electron has KE = 7.12 x 10-19 J = 4.45 eV. This is enough energy to cause an n = 1 to n = 3 transition, but not n = 1 to n = 4.
Problem 2.
a + b)
Don't worry about the fact that the signs are different from what we discussed in class today - your answers to the other parts are still consistent with the solutions below, and will be correct.
c) Highest energy is the 6 eV transition.
d) lowest is the 1 eV transition.
e) This is the reverse of c and d - longest wavelength will be the 1 eV transition since E = hc/lambda
f) shortest wavelength is with the 6 eV transition, which has the highest energy.
g) The difference in energy between these two energy levels is 5 eV. This translates into a photon of wavelength 248 nanometers.
h) Three are possible.
i) The electron has KE = 7.12 x 10-19 J = 4.45 eV. This is enough energy to cause an n = 1 to n = 3 transition, but not n = 1 to n = 4.
Problem 2.
1996B5.
a) Z = 102
b) K = 1/2*mv^2 = 8.42 MeV = 1.35 x 10-12 J
alpha particle mass = (4 u)*(1.66 x 10-27kg
v = (2*K/m)^1/2 = 2.02 x 107 m/s
c) A correct explanation includes a reference to:
1. a change in mass/mass defect being converted into KE fo the alpha particle
2. conservation of energy - binding energy of the nucleus is converted into KE of the alpha particle during the decay.
3. work-energy theorem - the electric force does work on the alpha particle as it moves away from the nucleus, thereby changing its KE.
Sunday, April 5, 2009
Nuke it - Mr. Weinberg has no Gas Stove
Chapter 30:
Q1. Isotopes of a single element have the same number of protons, so they act the same way in a chemical reaction as each other. They have different numbers of neutrons, so the mass of isotopes vary.
Q2 & 3.
a) Uranium-232, 92 protons, 140 neutrons
b) Nitrogen-13, 7 protons, 6 neutrons
c) Hydrogen, 1 proton, 0 neutrons
d) Strontium-82, 38 protons, 44 neutrons
e) Berkelium-247, 97 protons, 150 neutrons
Q10.
Gamma rays are photons or electromagnetic waves, so they have no mass and no charge. Alpha particles have a relatively large mass, a positive charge, and high energy. beta particles have a relatively small mass, a negative charge, and also high energy.
Weinberg Problems:
1.
a) 2.42 m/s
b) 0.642 Hz
c) 3.91 N
2.
a) 225 ohms
b) 75 ohms
3.
a) By the right hand rule, positive charges moving west in the Earth's magnetic field (directed North) will experience a downwards force, so the bottom part of the tube will become positive.
b) 200 m/s (pretty fast!)
4.
a) Green light has a wavelength between 500 and 550 nanometers, so any value in this range is a good estimate.
b) Using c = f*lambda and lambda = 525 nm, f = 5.7 x 1014 Hz
c) Frequency is the same, wavelength becomes 525 nm/ 1.4 = 375 nm.
d) The path difference is 2t, and the wave inverts twice upon reflection because the index of refraction of the reflecting surfaces is larger in both cases.
Thus 2t = m*lambda, and for the lowest nonzero thickness, t = 187.5 nm.
Q1. Isotopes of a single element have the same number of protons, so they act the same way in a chemical reaction as each other. They have different numbers of neutrons, so the mass of isotopes vary.
Q2 & 3.
a) Uranium-232, 92 protons, 140 neutrons
b) Nitrogen-13, 7 protons, 6 neutrons
c) Hydrogen, 1 proton, 0 neutrons
d) Strontium-82, 38 protons, 44 neutrons
e) Berkelium-247, 97 protons, 150 neutrons
Q10.
Gamma rays are photons or electromagnetic waves, so they have no mass and no charge. Alpha particles have a relatively large mass, a positive charge, and high energy. beta particles have a relatively small mass, a negative charge, and also high energy.
Weinberg Problems:
1.
a) 2.42 m/s
b) 0.642 Hz
c) 3.91 N
2.
a) 225 ohms
b) 75 ohms
3.
a) By the right hand rule, positive charges moving west in the Earth's magnetic field (directed North) will experience a downwards force, so the bottom part of the tube will become positive.
b) 200 m/s (pretty fast!)
4.
a) Green light has a wavelength between 500 and 550 nanometers, so any value in this range is a good estimate.
b) Using c = f*lambda and lambda = 525 nm, f = 5.7 x 1014 Hz
c) Frequency is the same, wavelength becomes 525 nm/ 1.4 = 375 nm.
d) The path difference is 2t, and the wave inverts twice upon reflection because the index of refraction of the reflecting surfaces is larger in both cases.
Thus 2t = m*lambda, and for the lowest nonzero thickness, t = 187.5 nm.
Thursday, April 2, 2009
Are you here from the Wave side, or the Particle side of the family?
1.
a) 2.8 eV
b) 1.7 x 1018 photons
c) 5.5 x 105 m/s
d) 1.3 x 10-9 m
2.
a) 6.38 x 10-10m
b)335 nm
c) The diagram consists of three transitions, 1.2, 3.7, and 4.9 eV.
3.
a) The best fit curve is shaped like a cosine curve.
b) 1.408 x 10-14 m
c) 4.71 x 10-20 kg*m/s
d) 8.08 x 10-14 J
4.
a) 1.33 x 1015 Hz
b) 276 nm (not visible), 497 nm, 621 nm
c) Because of the energy of the photons in the given range, transitions can occur from the ground state to either the 1st or 2nd excited states.
5.
a)122 nm
b)The energy is the difference between the ionization energy and the energy of the second photon.
13.6eV - 10.2 eV = 3.4 eV
c)8.2 x 1014Hz
d) 
1995B5.
Wednesday, April 1, 2009
I Emitted the Photon...YOU just didn't absorb it...
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
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