Friday, April 18, 2008

Solutions to Review Problems

Check back soon for updates:

Electrostatics Review Solutions

Waves/Interference Review Solutions:
1 D
2 A
3 C
4 C
5 B
6 A
7 D
8 E
9 E
10 E
11 C
12 D
13 A

Kinetic Theory/Thermo Solutions:
1 B
2 B
3 A
4 E
5 C
6 C
7 C
8 D
9 E
10 A
11 E
12 A
13 C
14 D
15 A
16 A

Wednesday, April 16, 2008

Quantum/Nuclear Solutions

1D
2A
3D
4C
5C
6B
7B
8B
9C
10C
11D
12A
13A
14D
15E
16A
17E
18A

Wednesday, April 9, 2008

Magnetism/EMI Solutions

1. a
2. b
3. c
4. a
5. a
6. c
7. b
8. e
9. b
10. b
11. a
12. a
13. c
14. a
15. e.
16. b
17. d
18. a
19. c
20. e

Tuesday, April 8, 2008

Useful Information

Arista.Weinberg.age_time_is_up!

Wednesday, April 2, 2008

Nuclear Radiation / Structure Solutions

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.

Tuesday, April 1, 2008

Solutions Galore - Atomic Energy levels

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) 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

3.
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.

4.
a) 6.38 x 10-10m
b)335 nm
c) The diagram consists of three transitions, 1.2, 3.7, and 4.9 eV.
7.
a) -3.4 eV
b) 5.44 x 10-27kg*m/s
c) 5.5 eV
d) 5.5 V