Nuclear & Particle Physics - Hydrogen

The spacing of the rotational energy levels for the hydrogen molecule H₂ is most nearly

A. 10⁻⁹ eV
B. 10⁻³ eV
C. 10 eV
D. 10 MeV
E. 100 MeV

(GR9277 #90)

Solution:

Rotational kinetic energy: E = ^L²∕_2I
Angular momentum: L² = l(l+1)ħ² with l = 0,1,2,⋯
Moment Inertia: I = mr²

For hydrogen atom:
r = 0.529 × 10⁻¹⁰ m
m =10⁻²⁷ kg (mass of proton)
→ I = 10⁻²⁷(0.529×10⁻¹⁰)² ≈ 10⁻⁴⁷ kgm²
ħ = ^h∕_2π = ^{6.63×10−34}∕ _2(3.14) ≈ 10⁻³⁴

For l = 0 → L = 0 → E = 0
For l = 1 → L = 2ħ² → E = ^ħ²∕_I

The spacing of the rotational energy levels:
∆E = ^ħ²∕_I  − 0 = ^(10−34)2∕ _10⁻⁴⁷ = 10⁻⁶⁸⁺⁴⁷ = 10⁻²¹ J

Convert to eV: 1 eV = 1.6 × 10⁻¹⁹ J
→ ∆E = ¹⁰⁻²¹∕ _{1.6 × 10⁻¹⁹} ≈ 10⁻³ eV

Answer: B

Quantum Mechanics - Infinite Potential Well

The figure above shows one of the possible energy eigenfunctions ψ(x) for a particle bouncing freely back and forth along the x-axis between impenetrable walls located at x = −a and x = +a. The potential energy equals zero for |x| > a. If the energy of the particle is 2 electron volts when it is in the quantum state associated with this eigenfunction, what is its energy when it is in the quantum state of lowest possible energy? 

A. 0 eV
B. 1/√2 eV
C. 1/2 eV
D. 1 eV
E. 2 eV

(GR8677 #90)

Solution:

Impenetrable walls = infinite potential walls.
The initial wavefunctions for the first four states in the system:

So, 

represents n = 2.

The energies for infinite potential walls:  E_n = n² E₁

Since E₂ = 2 eV → 2² E₁ = 2 → E₁ = 2/4 = 1/2 eV

Answer: C

Classical Mechanics - Oscillatory Motion

Two identical springs with spring constant k are connected to identical masses of mass M, as shown in the figures. The ratio of the period for the springs connected in parallel (Fig. 1) to the period for the springs connected in series (Fig.2) is

A. ½
B. 1/√2
C. 1
D. √2
E. 2

(GR0177 #90)

Solution:

Spring constants:

Series, k_s = ½ k
Parallel, k_p = 2k

Period:




Answer: A