Nuclear & Particle Physics - Hydrogen Spectrum

In the spectrum of Hydrogen, what is the ratio of the longest wavelength in the Lyman series (n_f  = 1) to the longest wavelength in the Balmer series (n_f  = 2)?

A. 5/27
B. 1/3
C. 4/9
D. 3/2
E. 3

(GR9677 #09)

Solution:

Rydberg formula:



λ_vac = the wavelength of the light emitted in vacuum
R_H = Rydberg constant for Hydrogen
n₁ and n₂ are integers such that n₁  n₂

For the longest wavelength take n₂ = ∞

For Lyman-radiation (n₂ = ∞ → n₁ = 1):

1/λ_L = R_H (1/1 − 0) = R_H

For Balmer-radiation (n₂ = ∞ → n₁ = 2):

1/λ_B = R_H (¼ − 0) = ¼R_H

The Ratio:

^λ_L/_λB = (¹/_RH)(^R_H/₄) = ¼ ≈ 5/27

Answer: A

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

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

Nuclear & Particle Physics - Beta Decay

When the beta decay of ⁶⁰Co nuclei is observed at low temperature in a magnetic field that aligns spins of the nuclei, it is found that the electrons are emitted preferentially in a direction opposite to the ⁶⁰Co spin direction. Which of the following invariances is violated by this decay

A. Gauge invariance
B. Time invariance
C. Translation invariance
D. Reflection invariance
E. Rotation invariance

(GR9677 #34)

Solution:

Wu experiment, 1957: beta decay of Cobalt-60 shows that electrons are emitted preferentially in a direction opposite to the ⁶⁰Co spin direction → Parity violation by the weak interaction.

A. Gauge invariance → related to conservation of charge
B. Time invariance → energy
C. Translation invariance → momentum
D. Reflection invariance → parity
E. Rotation invariance → angular momentum

Answer: D

Nuclear & Particle Physics - Helium

If a singly ionized Helium atom in an n = 4 state emits a photon of wavelength 470 nanometers, which of the following gives the approximate final energy level E_f  of the atom, and the n value, of n_f  this final state?

	Ef  (eV)	nf
A.	− 6.0	3
B.	− 6.0	2
C.	− 14	2
D.	− 14	1
E.	− 52	1

(GR9677 #40)

Solution:

E_photon = E_i  − E_f 

Helium: 2 electrons, 2 protons, 2 neutron
Singly ionized Helium, He⁺:1 electrons, 2 protons, 2 neutron
He⁺ → Hydrogen-like atom

Bohr's Equation for Hydrogen-like atom: E_n =  −13.6 Z²/n² eV

For Helium, Z = 2,
E_i = E_(n =4)  = − 13.6 (2)² /4² = − 13.6 /4 ≈ − 3.4 eV

E_photon  = hν = hc / λ 
with
h = 6.63 × 10⁻³⁴ Joule.second = 4.1 × 10⁻¹⁵ eV.second
c = 3 × 10⁸ m/s
λ  = 470 nm = 470 × 10⁻⁹ m = 4.7 × 10⁻⁷ m

E_photon = (4.1 × 10⁻¹⁵)(3 × 10⁸) / (4.7 × 10⁻⁷) ≈ 3 eV

E_f  = E_i − E_photon  
=  − 3.4 − 3
= − 6.4  eV

To find n_f :

n_f ² = −13.6 Z²/ E_f 
= −13.6 (2)²/ (− 6.4)
= 54.4/6.4 ≈ 9
n_f  = 3

Answer: A

Nuclear & Particle Physics - Photoelectric

Light of wavelength 500 nanometers is incident on sodium, with work function 2.28 electron volts. What is the maximum kinetic energy of the ejected photoelectrons?

A. 0.03 eV
B. 0.2 eV
C. 0.6 eV
D. 1.3 eV
E. 2.0 eV

(GR9677 #42)

Solution:

Einstein’s photoelectric equation:  |eV| = hv − W = ^hc/_λ − W

h = 4.1 × 10⁻¹⁵ eV.second
c = 3 × 10⁸ m/s
λ = 500 nm = 5 × 10⁻⁷ m
W = 2.28 eV

KE = [(4.1 × 10⁻¹⁵)(3 × 10⁸)/(5 × 10⁻⁷ )] − 2.28 
=  [(4.1 × 3 / 5)(10⁻¹⁵ × 10⁸ × 10⁷)] − 2.28
=  2.46 − 2.28 = 0.18 ≈  0.2 eV

Answer: B

Nuclear & Particle Physics - Positronium

Positronium is the bound state of an electron and a positron. Consider only the states of zero orbital angular momentum l = 0. The most probable decay product of any such state of positronium with spin zero (singlet is)

A. 0 photons
B. 1 photons
C. 2 photons
D. 3 photons
E. 4 photons

(GR9677 #53)

Solution:

The singlet state, s = 0 of Positronium is known as para-Positronium, decays preferentially into two gamma rays.

It can decay into any even number of photons (2, 4, 6, ...), but the probability quickly decreases with the number.

Thus, the most probable decay product of singlet state of positronium is 2 photons.

Answer: C

Nuclear & Particle Physics - Standard Model

According to the Standard Model of elementary particles, which of the following is NOT a composite object?

A. Muon
B. Pi-meson
C. Neutron
D. Deuteron
E. Alpha particle

(GR9677 #63)

Solution:

A.TRUE
Muon is a lepton. Leptons, along with quarks, are considered the fundamental particles.

B. FALSE
Pi-Meson consists of a quark and its antiparticle. Moreover, a pi-meson is a hadron. Hadrons interact with the strong-force, and all of them are composed of combinations of quarks. (The fundamental particles are classified as quarks and leptons.)

C. FALSE
A neutron is made up of 3 quarks.

D. FALSE
A deuteron consists of a proton and a neutron. (tritium is two neutrons and a proton, while regular Hydrogen is just an electron and proton).

E. FALSE
An alpha particle consists of electrons and protons and neutrons.

Answer: A

Nuclear & Particle Physics - Binding Energy

The binding energy of a heavy nucleus is about 7 million electron volts per nucleon, whereas the binding energy of a medium-weight nucleus is about 8 million electron volts per nucleon. Therefore, the total kinetic energy liberated when a heavy nucleus undergoes symmetric fission is most nearly

A. 1876 MeV
B. 938 MeV
C. 200 MeV
D. 8 MeV
E. 7 MeV

(GR9677 #64)

Solution:

KE = E_f  − E_i

Symmetric fission: the splitting of the nucleus into two fragments of approximately equal mass.

^AX  →  ^A₁Y  + ^A₂Z
with A₁ =  A₂ = A/2 and Y = Z  (for symmetric fission)

A E_i  → A₁ E_f 1 + A₂ E_f 2 = 2(A/2) E_f  = A E_f 

KE = A E_f  − A E_i  = A (8  − 7) MeV/nucleon = A MeV/nucleon

For heavy nucleus, A ≈ 200 nucleons. Example: ²³⁸U → A ≈ 238 nucleons

→ KE ≈ 200  MeV

Answer: C

Nuclear & Particle Physics - Bragg Diffraction

The longest wavelength X-ray that can undergo Bragg diffraction in a crystal for a given family of planes of spacing d is

A. d/4
B. d/2
C. d
D. 2d
E. 4d

(GR9277 #02)

Solution:

Bragg’s law: 2d sin θ = nλ

Maximum → sin θ = 1

and n = 1 (1st order)

λ = 2d

Answer: D

Nuclear & Particle Physics - X-rays

The ratio of the energies of the K characteristic X-rays of carbon (Z = 6) to those of magnesium (Z = 12) is most nearly

A. 1/4
B. 1/2
C. 1
D. 2
E. 4

(GR9277 #03)

Solution:
Moseley’s law:

 

K-series refers to a transition from some outer state, n_i to the inner-most shell, n_f = 1.
(The order from inner to outer → K, L, M, N).

→ E ≈ (Z − 1)²

E_carbon = (6 − 1)² = 25
E_magnesium = (12 − 1)² = 121

The ratio = 25/121 ≈ 1/4

Answer: A

Nuclear & Particle Physics - Bonding Mechanism

Solid Argon is held together by which of the following bonding mechanism?

A. Ionic bond only
B. Covalent bond only
C. Partly covalent and partly ionic bond
D. Metallic bond
E. Van der Waals bond

(GR9277 #24)

Solution:

Solid argon →  a noble gas full outermost electron shell

(A) FALSE
Ionic bond only → giving/receiving electron

(B) FALSE
Covalent bond only → sharing electron

(C) FALSE
Partly covalent and partly ionic bond

(D) FALSE
Metallic bond → a positive charge/nucleus swimming in a sea of free electron → FALSE

(E) TRUE
Van der Waals bond:

• occur between atoms or molecules of the same type
• occur due to variances in charge distribution in the atom

Answer: E 

Nuclear & Particle Physics - Cosmic Rays

In experiments located deep underground the two types of cosmic rays that most commonly reach the experimental apparatus are

A. alpha particles and neutrons
B. protons and electrons
C. iron nuclei and carbon nuclei
D. muons and neutrinos
E. positrons and electrons

(GR9277 #25)

Solution:

Since the experiment is located deep underground, the particles must be essentially massless and able to pass through or do not interact with matter easily → muons and neutrinos.

Muon:

• "heavy" electron (200 times the mass of the electrons)
• highly penetrating

Neutrinos:

• almost no mass
• travel close to the speed of light
• weakly interact with other

Answer: D



Notes:

• Alpha particles and neutrons have high kinetic energy but very short penetrating depth because they are quite massive, can't go through the Earth.
• Protons and electrons don’t penetrate much as they interact easily with matter. Protons are also pretty massive.
• Iron and carbon nuclei are very heavy and interact very easily with matter.
• Positrons and electrons have the same mass and are highly interacting with matter. Positron also has a very short life span.

Nuclear & Particle Physics - Radioactive

A radioactive nucleus decays, with the activity shown in the graph above. The half-life of the nucleus is

A. 2 min
B. 7 min
C. 11 min
D. 18 min
E. 23 min

(GR9277 #26)

Solution:

From the graph: at t = 0, N ≈ 6 × 10³

Half life, N_½ = 3 × 10³ → t = 7

Answer: B

Nuclear & Particle Physics - Positronium

Given that the binding energy of the hydrogen atom ground state is E₀ = 13.6 eV, the binding energy of n = 2 state of positronium (positron-electron system) is

A. 8E₀
B. 4E₀
C. E₀
D. E₀/4
E. E₀/8

(GR9277 #30)

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 n = 2,

E_2(Ps) = ½ × ( − 13.6 / 2²)  = − E₀/8

Answer: E

Nuclear & Particle Physics - Spectroscopic Notation

In a ³S state of the helium atom, the possible values of the total electronic angular momentum quantum number are

A. 0 only
B. 1 only
C. 0 and 1 only
D. 0, 1/2, and 1
E. 0, 1, and 2

(GR9277 #31)

Solution:

Spectroscopic notation: N^2s+1 L_j=l+s

• N = the principal quantum number and will often be omitted
• s = the total spin quantum number
• L = the orbital angular momentum quantum number, l but is written as S, P, D, F, … for l = 0, 1, 2, 3,⋯
• j = the total angular momentum quantum number

For ³S 
S → l = 0
3 → 3 = 2s + 1 → s = 1
j = l + s = 0 + 1 = 1

Answer: B 

Nuclear & Particle Physics - Muon

Two horizontal scintillation counters are located near the Earth’s surface. One is 3.0 meters directly above the other. Of the following, which is the largest scintillator resolving time that can be used to distinguish downward-going relativistic muons from upward-going relativistic muons using the relative time of the scintillator signals?

A. 1 picosecond
B. 1 nanosecond
C. 1 microsecond
D. 1 millisecond
E. 1 second

(GR9277 #49)

Solution:

Muons can travel with maximum speed near the speed of light:
v = 3 × 10⁸ m/s

Given s = 3 m,

t = s/v = 3/(3 × 10⁸) = 10⁻⁸

Micro = 10⁻⁶
Nano = 10⁻⁹
Pico = 10⁻¹²

10⁻⁸ second is close to 1 nanosecond.

Answer: B

Nuclear & Particle Physics - Electron Configuration

The ground state configuration of a neutral sodium atom (Z = 11) is
   
A. 1s² 2s² 2p⁵ 3s²
B. 1s² 2s³ 2p⁶
C. 1s² 2s² 2p⁶ 3s   
D. 1s² 2s² 2p⁶ 3p
E. 1s² 2s² 2p⁵

(GR9277 #58)

Solution:

Z = 11 → 1s² 2s² 2p⁶ 3s¹

Answer: C


Note:

Electron configuration:

	l = 0	l = 1	l = 2	l = 3	# of electrons
n = 1	1s2				2
n = 2	2s2	2p6			8
n = 3	3s2	3p6	3d10		18
n = 4	4s2	4p6	4d10	4f14	32