Quantum Mechanics - Harmonic Oscillator

The energy levels for the one-dimensional harmonic oscillator are hν(n + ½), n = 0,1,2,⋯ How will the energy levels for the potential shown in the graph above differ from those for the harmonic oscillator?

A. The term ¹⁄₂ will be changed to ³⁄₂
B. The energy of each level will be doubled.
C. The energy of each level will be halved.
D. Only those for even values of n will be present.
E. Only those for odd values of n will be present.

(GR9277 #89)

Solution:

See GR9677 #98

Answer: E

Quantum Mechanics - Wave Function

A system containing two identical particles is described by a wave function of the form

Where x₁ and x₂ represent the spatial coordinates of the particles and α and β represent all the quantum numbers, including spin, of the states that they occupy. The particles might be 

A. Electrons
B. Positrons
C. Protons
D. Neutrons
E. Deuterons

(GR8677 #89)

Solution:

Wave function: 

Symmetric function (+ sign):

→ obey Bose-Einstein statistics → bosons

Anti-symmetric function (−sign):
→ obey Fermi-Dirac statistics → fermion

Electrons and positron (anti-elecron) → fermion

Protons and neutron → fermionic hadrons

• composite particles (hadron)
• composed of 3 fermionic quarks

Deuterons → bosons

• nucleus of deuterium, an isotope of hydrogen
• composed of a neutron (spin 1/2) and a proton (spin 1/2), total spin = 1

Answer: E

Electromagnetism - Particle's Trajectory

A particle with charge q and momentum p is moving in the horizontal plane under the action of a uniform vertical magnetic field of magnitude B. Measurements are made of the particle's trajectory to determine the “sagitta” s and half-chord length l, as shown in the figure above. Which of the following expressions gives the particle's momentum in terms of q, B, s, and l? (Assume s ≪ l).

A. qBs²/2l
B. qBs²/l
C. qBl/s
D. qBl²/2s
E. qBl²/8s

(GR9677 #89)

Solution:



Answer: D

Classical Mechanics - Rotational Motion

A child is standing on the edge of a merry-go-around that has the shape of a solid disk, as shown in the figure. The mass of the child is 40 kilograms. The merry-go-around has a mass 200 kilograms and a radius of 2.5 meters and it is rotating with an angular velocity of 2.0 radians per second. The child then walk slowly toward the center of the merry-go-around. What will the final angular velocity of the merry-go-around when the child reaches the center? (The size of the child can be neglected).

A. 2.0 rad/s
B. 2.2 rad/s
C. 2.4 rad/s
D. 2.6 rad/s
E. 2.8 rad/s

(GR0177 #89)

Solution:

Angular Momentum: L = Iω

Conservation of Angular Momentum:

∑ L_i = ∑ L_f

(I_m + I_c)ω = (I_m' + I_c')ω' 
ω' = ω(I_m + I_c)/(I_m' + I_c')

Moment inertia of the uniform disk is the same before and after the child moves to the center:

I_m = I_m' = I_m = ½ m_mr² = ½ (200)(2.5)²
ω = 2 rad/s

Initally, moment inertia of the child:

I_c = m_cr² = (40)(2.5)²

When the child reaches the center, r = 0:

I_c' = 0

Thus,

ω' = ω(I_m + I_c)/(I_m' + I_c')
= (2) [½(200)(2.5)² + (40)(2.5)²] / [½ (200)(2.5)²]
= (2) [½(200) + (40)] / [½ (200)]
= 280/100
= 2.8 rad/s

Answer: E