Quantum Mechanics - Spherical Harmonics

A diatomic molecules is initially in the state Ψ(Θ, Φ) = (5Y₁¹ + 3Y₅¹ + 2Y₅⁻¹) / (38)^½ where Y_l^m is a spherical harmonics. If measurements are made of the total angular momentum quantum number l and of azimuthal angular momentum quantum number m, what is the probability of obtaining the results l = 5?

A. 36/1444
B. 9/38
C. 13/38
D. 5/(38)^½
E. 34/38

(GR9677 #33)

Solution:

l = 5 → Y₅¹ and Y₅⁻¹

Probability, P = ∑_i |c_i|² 

P = |3/√38|² + |2/√38|² =  9/38 + 4/38 =  13/38

Answer: C

Electromagnetism - Kirchoff's Law

See Question #32. The voltage across resistor R₄ is

A. 0.4 V
B. 0.6 V
C. 1.2 V
D. 1.5 V
E. 3.0 V

(GR9277 #33)

Solution:

Parallel: V_t = V₁ = V₂ = V₃ = ...

Series: V_t = V₁ + V₂ + V₃ + ...

V₄ = V₃ = V₃₄
V₃₄ + V₅ = V₂ 
V₂ + V₁ = V_t 

Or, using Kirchoff's Law:
Loop 1: V_t  = V₁ + V₂  
Loop 2: V_t  = V₁ + V₃₄ + V₅

From Question #32:
R₃₄ = 20
R₃₄₅ = 50
R₂₃₄₅ = 25

R_t = R₁ + R₂₃₄₅ = 50 + 25 = 75
I_t = V_t / R_t = 3/75 = I₁
V₁ =  I₁ R₁ = (3/75) × 50 = 2

Thus, V₃₄ + V₅ = V_t −V₁ = 3 − 2 = 1

V₃₄ must be less than 1
Only (A) and (B) could be TRUE

Since R₄ = R₅ but R₄ is in parallel with R₃ → V₄ must be less than V₅ 

Answer: A

Complete Calculation:

I₁ =  I₂ + I₃₄₅

R₃₄₅ = 50 = R₂
I₃₄₅ = I₂ =  I₁ / 2 = (3/75) / 2 = 3/150

R₃₄ and R₅ are in series
I₃₄₅ = I₃₄ = I₅
V₄ = V₃₄ = I₃₄ R₃₄ = (3/150) × 20 = 60/150 = 0.4

Nuclear & Particle Physics - Photoelectric

Questions 31-33  refer to the apparatus used to study the photoelectric effect (see GR8677 #31).
The quantity W in the photoelectric equation is the

1. Energy difference between the two lowest electron orbits in the atoms of the photocathode
2. Total light energy absorbed by the photocathode during the measurement
3. Minimum energy a photon must have in order to be absorbed by the photocathode
4. Minimum energy required to free an electron from its binding to the cathode materials
5. Average energy of all electrons in the photocathod

(GR8677 #33)

Solution:

Einstein photoelectric equation: |eV| = hv −W

eV = kinetic energy of the electron as it accelerates through the medium between the cathode and collector
W = work function; energy to free the electron from the metal
hv = energy given by the light source.

Answer: D

Special Relativity - Relativistic Speed

If a charged pion that decay in 10⁻⁸ second in its own rest frame is to travel 30 meters in the laboratory before decaying, the pion’s speed must be most nearly

A. 0.43 × 10⁸ m/s
B. 2.84 × 10⁸ m/s
C. 2.90 × 10⁸ m/s
D. 2.98 × 10⁸ m/s
E. 3.00 × 10⁸ m/s

(GR0177 #33)

Solution:

v = v₀/γ 
with γ = 1/(1 − ^v2/_c²)^½ 

Given:
t₀ = 10⁻⁸ second 
L₀ = 30 meters
v₀ = L₀/t₀ = 30 × 10⁸ = 10(3 × 10⁸) = 10c

v = v₀(1 − ^v2/_c²)^½
v² = v₀²(1 − ^v2/_c²) = (v₀² − v²v₀²/c²) = 100c² − 100v²
101v² = 100c²
v = (100/101)c = 0.99c = 0.99(3 × 10⁸) ≈ 2.98 × 10⁸ m/s

Answer: D