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 - 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

Special Relativity - Doppler Effect

The Lyman alpha spectral line of Hydrogen (λ = 122 nanometers) differs by 1.8 × 10⁻¹² meter in spectra taken at opposite ends of the Sun’s equator. What is the speed of a particle on the equator due to the Sun’s rotation, in kilometers per second?

A. 0.22
B. 2.2
C. 22
D. 220
E. 2200

(GR9677 #60)

Solution:

The problem deals with v ≪ c since the  answer is in km/s.

For v ≪ c, Redshift parameter, z = Δλ/λ ≈ v/c

v =  cΔλ/λ

Given:
Δλ = 1.8 × 10⁻¹² m 
λ = 122 nm = 1.22 × 10⁻⁷ m
c = 3 × 10⁸ m/s

v = (3 × 10⁸)(1.8 × 10⁻¹²)/(1.22 × 10⁻⁷) 

= (5.4/1.22) × 10³ m/s  
= (5.4/1.22) km/s ≈ 2.2 km/s

Answer: B

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

Nuclear & Particle Physics - Hydrogen

The energy levels of the hydrogen atom are given in terms of the principal quantum number n and a positive constant A by the expression 

A. A(n + ½)
B. A(1 − n²)
C. A(¼ + 1/n)
D. An²
E. −A/n²

(GR8677 #19 )

Solution:

Energy level of hydrogen atom: E_n = −13.6/n² eV

Answer: E

Nuclear & Particle Physics - Selection Rules

A transition in which one photon is radiated by the electron in a hydrogen atom when the electron's wave function changes from ψ₁ to ψ₂ is forbidden if ψ₁ and ψ₂

A. have opposite parity
B. are orthogonal to each other
C. are zero at the center of the atomic nucleus
D. are both spherically symmetrical
E. are associated with different angular momenta

(GR8677 #48)

Solution:

Selection rules:

1.	Principal quantum number	:	∆n = anything
2.	Orbital angular momentum	:	∆l = ±1
3.	Magnetic quantum number	:	∆ml = 0, ±1
4.	Spin	:	∆s = 0
5.	Total angular momentum	:	∆j = 0, ±1, but j = 0 ↛j = 0

A. FALSE
It’s not related to the selection rules.

B. FALSE
In any transition, eigenstates should always be mutually orthogonal.

C. FALSE.
Eigenstates zero at the center → l 0 could change, for example from 3 to 2. This is allowed.

D. TRUE.
If both initial and final states have spherically symmetrical wave functions, then they have the same angular momentum. l = 0 → l = 0 is forbidden.

E. FALSE.
The selection rules require l to change.

Answer: D

Nuclear & Particle Physics - The Sun’s Energy

The primary source of the sun’s energy is a series of thermonuclear reactions in which the energy produces is c² times the mass difference between

A. two hydrogen atoms and one helium atom
B. four hydrogen atoms and one helium atom
C. six hydrogen atoms and two helium atoms
D. three helium atoms and one carbon atom
E. two hydrogen atoms plus two helium atoms and one carbon atom

(GR0177 #19)

Solution:

In the Sun’s core, the protons of the hydrogen atoms (which largely make up the Sun) collide into each other and stick together or “fuse” to create helium nuclei.

4H → He + energy

It takes 4H nuclei to create a He nucleus due to the conservation of atomic mass.

Hydrogen atomic mass = 1 proton = 1.
Helium atomic mass = 2 protons + 2 neutrons = 4.

Answer: B

Nuclear & Particle Physics - Hydrogen Spectrum

In the hydrogen spectrum, the ratio of the wavelength for Lyman-radiation  (n = 2  to  n = 1) to Balmer-radiation (n = 3  to  n = 2)  is

A. 5/48
B. 5/27
C. 1/3
D. 3
E. 27/5

(GR0177 #21)

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

For Lyman-radiation (n = 2 → n = 1):



For Balmer-radiation (n = 3 → n = 2):



The Ratio:



Answer: B

Special Relativity - Doppler Effect

The ultraviolet Lyman alpha line of hydrogen with wavelength 121.5 nanometers is emitted by an astronomical object. An observer on earth measures the wavelength of the light received from the object to be 607.5 nanometers. The observer can conclude that the object is moving with radial velocity of

A. m/s toward Earth
B. m/s toward Earth
C. m/s away from Earth
D. m/s away from Earth
E. m/s away from Earth

(GR0177 #71)

Solution:

Given:
λ₀ = 121.5 nm
λ = 607.5 nm

λ  λ₀ → the object is moving away (receding)

(A) and (B) are FALSE.

(E) is FALSE since v is larger than c.

Doppler Effect for light:


with

and
+ sign = approaching
− sign = receding

Since the object and the source are receding:

λ = λ₀(1 + β)^1/2/(1 − β)^1/2
λ/λ₀ = (1 + β)^1/2/(1 − β)^1/2
607.5/121.5 = 5 = (1 + β)^1/2/(1 − β)^1/2
25(1 − β) = (1 + β)
24 = 26β
β = 24/26 = 12/13 = v_source/c
v_source = (12/13)c = (12/13)(3 × 10⁸) = (36/13) × 10⁸ = 2.76 × 10⁸ m/s

Answer: D

Nuclear & Particle Physics - Selection Rules

An energy-level diagram of the n = 1 and n = 2 levels of atomic hydrogen (including the effect of spin-orbit coupling and relativity) is shown in the figure. Three transitions are labeled A, B, and C. Which of the transitions will be possible electric-dipole transition?

A. B only
B. C only
C. A and C only
D. B and C only
E. A, B, and C

(GR0177 #84)

Solution:

Selection rules:

1.	Principal quantum number	:	∆n = anything
2.	Orbital angular momentum	:	∆l = ±1
3.	Magnetic quantum number	:	∆ml = 0, ±1
4.	Spin	:	∆s = 0
5.	Total angular momentum	:	∆j = 0, ±1, but j = 0 ↛j = 0

Transition A:  
l = 0 to l = 0 → ∆l = 0 → NOT ALLOWED

Transition B:
l = 1 to l = 0 → ∆l = −1 → ALLOWED
j = 3/2 to j = 1/2 → ∆j = −1 → ALLOWED

Transition C:
 j = 1/2 to j = 1/2 → ∆j = 0 → ALLOWED

Answer: D

Quantum Mechanics - Bohr Radius

The solution to the Schrodinger equation for the ground state of hydrogen is



where a₀ is the Bohr radius and r is the distance from the origin. Which of the following is the most probable value for r?

A. 0
B. a₀ / 2
C. a₀
D. 2a₀
E. ∞

(GR0177 #93)

Solution:

Probability



The most probable value of r corresponds to the peak of the plot of P(r) versus r.
The slope of the curve at this point is zero.





For  

r = a₀

Answer: C