Nuclear & Particle Physics - Internal Conversion

Internal conversion is the process whereby an excited nucleus transfers its energy directly to one of the most tightly bound atomic electrons, causing the electron to be ejected from the atom and leaving the atom in an excited state. The most probable process after an internal conversion electron is ejected from an atom with a high atomic number is that the

1. atom returns to its ground state through inelastic collisions with either atoms
2. atom emits one or several X-rays
3. nucleus emits a gamma-ray
4. nucleus emits an electron
5. nucleus emits a positron

(GR9677 #10)

Solution:

Electron transitions in atom (internal conversion) = X-ray production
→ An orbital electron is absorbed and ejected along with an X-ray

compared to:

Nuclear transitions = Gamma, γ Ray production

→ The excited nucleus jumps to a lower level and emits a photon γ

Answer: B

Note: 

1. (C), (D), and (E) are products of radioactive decay which are results of unstable nuclei.

2. In internal conversion:

• For low atomic number, it will produce the Auger effect and ionize the outside electron.
• For high atomic number, it will only emit X-rays.

Nuclear & Particle Physics - Stern-Gerlach Experiment

A beam of neutral hydrogen atoms in their ground state is moving into the plane of this page and passes through a region of a strong inhomogeneous magnetic field that is directed upward in the plane of the page. After the beam passes through this field, a detector would find that it has been

A. deflected upward
B. deflected to the right
C. undeviated
D. split vertically into two beams
E. split horizontally into three beams

(GR9677 #11)

Solution:

The Stern–Gerlach experiment → spin discovery

A beam of neutral atom passes through inhomogeneous magnetic field will split vertically into 2 beams representing spin-up and spin-down electrons.

Answer: D

Nuclear & Particle Physics - Positronium

The ground-state energy of positronium is most nearly equal to

A. − 27.2 eV
B. − 13.6 eV
C. − 6.8 eV
D. − 3.4 eV
E. 13.6 eV

(GR9677 #12)

Solution:

Energy levels of Positronium is half those of Hydrogen (See GR8677 #99)

E_n(H)  = − 13.6 / n²
E_n(Ps) = ½ E_n(H) 

For the ground-state n = 1 → E_Ps = − ½ × 13.6 eV = − 6.8 eV

Answer: C

Thermal Physics - Power

A 100-watt electric heater element is placed in a pan containing one liter of water. Although the heating element is on for a long time, the water, though close to boiling does not boil. When the heating element is removed, approximately how long will it take the water to cool by 1 degree Celsius? (Assume that the specific heat for water is 4.2 kJ/kg^oC)

A. 20 s
B. 40 s
C. 60 s
D. 130 s
E. 200 s

(GR9677 #13)

Solution:

P = 100 W
V = 1 L = 1 m³ → m = 1 kg (STP)
ΔT = 1^oC
c = 4.2 kJ/kg^oC

Q = mcΔT = Pt
1 × 4200 × 1 = 100t
t = 42 s

Answer: B

Thermal Physics - Specific Heat

Two identical 1 kg blocks of copper metal one initially at a temperature T₁ = 0^oC and the other initially at a temperature T₂ = 100^oC are enclosed in a perfectly insulating container. The two blocks are initially separated. When the blocks are placed in contact, they come to equilibrium at a final temperature T_f. The amount of heat exchanged between the two blocks in this process is equal to which of the following? (The specific heat of copper metal is equal to 0.1 kilocalorie/kg^oK)

A. 50 Kcal
B. 25 Kcal
C. 10 Kcal
D. 5 Kcal
E. 1 Kcal

(GR9677 #14)

Solution:

Q = mcΔT
|Q|_gain = |Q|_lost 
m₁c₁ΔT₁  = m₂c₂ΔT₂

m₁ = m₂ = 1 kg
c₁ = c₂ = c_copper = 0.1 kcal/kg^oK

T₁ = 0^oC = 273^oK

T₂ = 100^oC = 373^oK

ΔT₁ = ΔT₂
T_f   − T₁ = T₂ − T_f 
T_f  = (T₂ + T₁)/2 = (100 + 0)/2 = 50^oC = 323^oK

|Q|_gain = |Q|_lost  = m₁c₁ΔT₁  = 1 × 0.1 × (323 − 273) = 5kcal

Answer: D

Thermal Physics - Isothermal

Suppose one mole of an ideal gas undergoes the reversible cycle ABCA shown in the P-V diagram above, where AB is an isotherm. The molar heat capacities are C_p at constant pressure and C_v at constant volume. The net heat added to the gas during the cycle is equal to

A. RT_h (V₂/V₁)
B. −C_p(T_h − T_c)
C. C_p(T_h − T_c)
D. RT_h ln (V₂/V₁) − C_p(T_h − T_c)
E. RT_h ln (V₂/V₁) − R(T_h − T_c)

(GR9677 #15)

Solution:

AB Isotherm → T Constant = T_h

Ideal Gas: PV = nRT 
P = nRT / V

n = 1 mole,

W_AB = _V1∫^V₂ P dV = RT_h V1∫^V₂ (1/V) dV  = RT_h ln (V₂/V₁)

BC Isobaric → P constant = P₂

W_BC = _V2∫^V₁ P dV = P₂ (V₁− V₂) =  P₂V₁ −  P₂V₂

From the diagram:

P₂V₁ = nRT_c 

P₂V₂ = nRT_h

W_BC = nR(T_c − T_h)  = R(T_c − T_h)

W_CA = 0  since V constant

Total W = W_AB + W_BC = RT_h ln (V₂/V₁) + R(T_c − T_h)

or

W = RT_h ln (V₂/V₁) − R(T_h − T_c)

Answer: E

Thermal Physics - Mean Free Path

The mean free path for the molecules of a gas is approximately given by 1/ησ, where η is the number density and σ is the collision cross section. The mean free path for air molecules at room conditions is approximately

A. 10⁻⁴ m
B. 10⁻⁷ m
C. 10⁻¹⁰ m
D. 10⁻¹³ m
E. 10⁻¹⁶ m

(GR9677 #16)

Solution:

Mean free path = 1/ησ

Number density, η = N/V  
Cross section area, σ = πr² 

For Ideal Gas: PV = NkT

1/ησ = V / Nπr²  = NkT / PNπr² = kT / Pπr²   

k = 1.38 × 10⁻²³ Joule/K
Radius of atom is in order of Angstrom: 10⁻¹⁰ m
P (STP) = 1 atm = 10⁵ Newton/meter²
T (STP) = 0 °C = 32 °F = 273.15 K ≈ 2 × 10² K

kT / Pπr² = (1.38 × 10⁻²³ × 2 ×10²) / (π ×10⁵ × 10⁻²⁰) 
= (1.38 × 2 / π) 10^{−23+2−5+20} ≈ 10⁻⁷

Answer: B

Quantum Mechanics - Probability

The wave function for a particle constrained to move in one dimension is shown in the graph (Ψ = 0 for x ≤  0 and x ≥ 5). What is the probability that the particle would be found between x = 2 and x = 4?

A. 17/64
B. 25/64
C. 5/8
D. v(5/8)
E. 13/16

(GR9677 #17)

Solution:

Probability, P ∼ ψ²

Probability to find the particle between x = 2 and x = 4 (unnormalized probability):
ψ² = 2²  + 3² = 13

Total probability (normalized probability):
ψ² = 1²  +  1²  + 2²  + 3²  +  1² = 16

P = unnormalized probability / normalized probability = 13/16

Answer: E

Quantum Mechanics - Potential Wall

Consider a potential of the form

V(x) = 0, x ≤ a
V(x) = V_0, a < x < b
V(x) = 0, x ≥ b

As shown in the figure above. Which of the following wave functions is possible for a particle incident from the left with energy E < V₀.

(GR9677 #18)

Solution:

A. Classic not QM potential → FALSE
B. No decrease in amplitude → FALSE
C. Decay exponentially inside the wall, decrease amplitude (fits E < V₀) →  TRUE

D. QM Oscillator harmonics → FALSE
E. Cosine wave function not QM potential → FALSE

Answer: C

Nuclear & Particle Physics - Alpha/Rutherford Scattering

When alpha particles are directed onto atoms in a thin metal foil, some make very close collisions with the nuclei of the atoms and are scattered at large angles. If an alpha particles with an initial kinetic energy of 5 MeV happens to be scattered through an angle of 180^o, which of the following must have been its distance of the closest approach to the scattering nucleus? (Assume that the metal foil is made of silver, with Z = 50.)

A. 1.22 × 50^1/3 fm
B. 2.9 × 10⁻¹⁴ m
C. 1.0 × 10⁻¹² m
D. 3.0 × 10⁻⁸ m
E. 1.7 × 10⁻⁷ m

(GR9677 #19)

Solution:

"When alpha particles are directed onto atoms in a thin metal foil, some make very close collisions with the nuclei of the atoms and are scattered at large angles."→ Rutherford Scattering: the discovery of nucleus.

Rutherford estimated the radius of a silver nucleus to be 2 × 10⁻¹⁴ m, by observing the angular dependence of alpha-particle scattering (source).

Answer: B

Calculation:

Conservation of Energy: U = T

U = kq_α q_s / r = T
r =  kq_α q_s / T

Given:
T = 5 MeV = 5 × 10⁶ eV
Z_s = 50 → q_s = Z_αe = 50e
Z_α = 2 → q_α = Z_αe = 2e 

k = 1/4πɛ₀ = 1/(4 × 3.14 × 8.85 × 10⁻¹²)  ≈ 10¹⁰  Nm²/C²

r = (10¹⁰ Nm²/C²× 50e × 2e) / (5 × 10⁶ eV)
= 2 × 10⁵ e Nm²/VC² 

With e = 1.60 × 10⁻¹⁹ C

1 Volt = 1 Nm/C

r = 2 × 10⁵ × 1.60 × 10⁻¹⁹ m ≈ 3 × 10⁻¹⁴ m