Condensed Matter - Semiconductor

In an n-type semiconductor, which of the following is true of impurity atoms?

1. They accept electrons from the filled valence band into empty energy levels just above the valence band.
2. They accept electrons from the filled valence band into empty energy levels just below the valence band.
3. They accept electrons from the conduction band into empty energy levels just below the conduction band.
4. They donate electrons to the filled valence band from donor levels just above the valence band.
5. They donate electrons to the conduction band from filled donor levels just below the conduction band.

(GR9677 #78)

Solution:

Semiconductor:

n-type:

• n for negative-charge
• atoms have extra electron
• donors to the conduction band

p-type:

• p for positive-charge (holes)
• atoms have missing electrons
• acceptors in the valence band

Source: physics.udel.edu

Answer: E 

Classical Mechanics - Rotational Motion

One ice skater of mass m moves with speed 2v, while another of the same mass m moves with speed v toward the left, as shown in Figure I. Their paths are separated by a distance b. At t = 0, when they are both at x = 0, they grasp a pole of length b and negligible mass. For t > 0, consider the system as a rigid body of two masses m separated by distance b, as shown in Figure II. Which of the following is the correct formula for the motion after t = 0 of the skater initially at y = b/2?

A.	x = 2vt	y = b/2
B.	x = vt + 0.5b sin (3vt/b)	y = 0.5b cos (3vt/b)
C.	x = 0.5vt + 0.5b sin (3vt/b)	y = 0.5b cos (3vt/b)
D.	x = vt + 0.5b sin (6vt/b)	y = 0.5b cos (6vt/b)
E.	x = 0.5vt + 0.5b sin (6vt/b)	y = 0.5b cos (6vt/b)

(GR9277 #78)

Solution:
x = x_{translational} + x_rotational

To find  x_trans:
Conservation of linear momentum: m₁v₁ + m₂v₂ = (m₁ + m₂)v'
m(2v) + m(−v) = 2mv'
→ v' = 0.5v
→ x_trans = v't = 0.5vt
→ A, B and E are FALSE.

Check answer C. x = 0.5vt + 0.5b sin (3vt/b)
→ x_rot = a sin ωt = 0.5b sin (3vt/b) → ω = 3v/b

To check if ω = 3v/b:

At t = 0,
L = L₁ + L₂ = r₁m₁v₁ + r₂m₂v₂ = ¹/₂bm2v + ¹/₂bmv = ³/₂bmv

At t > 0,  
L' = Iω
Moment inertia of the system: I = ½mb²
→ L' = ½mb²ω

Conservation of angular momentum:  L = L' 
→ ³/₂bmv = ¹/₂mb²ω → ω = 3v/b

Answer: C

Electromagnetism - Energy

A system consists of two charged particles of equal mass. Initially the particles are far apart, have zero potential energy, and one particle has nonzero speed. If radiation is neglected, which of the following is true of the total energy of the system?

1. It zero and remains zero
2. It is negative and constant
3. It is positive and constant
4. It is constant, but the sign cannot be determined unless the initial velocities of both particles are known.
5. It cannot be a constant of the motion because the particles exert force on each other.

(GR8677 #78)

Solution:

Particles have zero potential energy → PE = 0

One particle has nonzero speed → v 0 → KE 0

→ Total energy: E = PE + KE = 0 + (KE 0) → E  0

No radiation → no energy loss → Energy constant.

→ Energy is positive and constant.

Answer: C

Nuclear & Particle Physics - Conservation Laws

The muon decays with a characteristic lifetime of about 10⁻⁶ second into an electron, a muon neutrino, and an electron antineutrino. The muon is forbidden from decaying into an electron and just a single neutrino by the law of conservation of

A. charge
B. mass
C. energy and momentum
D. baryon number
E. lepton number

(GR0177 #78)

Solution:

If  μ → e⁻ + ν

A. FALSE. Conservation of charge: − 1 → − 1 + 0 is not violated.

B. FALSE. Conservation of mass cannot be violated in any decay/interactions.

C. FALSE. Conservation of energy and momentum cannot be violated in any decay/interactions.

D. FALSE. Muon is a lepton. It follows the Conservation of Lepton Number, not Baryon Number.

E. TRUE. Lepton Numbers: 1 → 1 + 1 is violated.

Answer: E