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

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 - Zeeman Effect

The emission spectrum of an atomic gas in a magnetic field differs from that of the gas in the absence of a magnetic field. Which of the following is true of the phenomenon?

A. It is called the Stern-Gerlach effect
B. It is called the Stark effect
C. It is due primarily to the nuclear magnetic moment of the atoms
D. The number of emission lines observed for the gas in a magnetic field is always twice the number observed in the absence of a magnetic field.
E. The number of emission lines observed for the gas in a magnetic field is either greater than or equal to the number observed in the absence of a magnetic field

(GR8677 #82)

Solution:

Zeeman Effect: the splitting of a spectral line into several components in the presence of a static magnetic field.→ A and B are FALSE.

In most atoms, there exist several electron configurations with the same energy (degeneracy), so that transitions between these configurations and another correspond to a single spectral line.

The presence of a magnetic field breaks this degeneracy, since the magnetic field interacts differently with electrons with different quantum numbers, slightly modifying their energies. → C is FALSE.

The result is that, where there were several configurations with the same energy, they now have different energies, giving rise to several very close spectral lines.

image:  physics.cornell.edu (click image to enlarge)

→ E is TRUE as opposed to D

Answer: E

Nuclear & Particle Physics - EM Spectrum

A spectral line is produced by a gas that is sufficiently dense that the mean time between atomic collisions is much shorter than the mean lives of the atomic states responsible for the line. Compared with the same line produced by a low-density gas, the line produced by the higher-density gas will appear

A. The same
B. More highly polarized
C. Broader
D. Shifted toward the blue end of the spectrum
E. Split into a doublet

(GR8677 #83)

Solution:

Spectral lines = bright or dark lines correspond to the emission or absorption of light of a single wavelength.

High-density gas:

• Atoms experience frequent collisions and many other mechanisms that cause spectral lines to lose their sharpness.
• It produces a continuum spectrum = weaker, less distinct or broader spectral lines.

Low-density gas:

• Atoms do not experience many collisions (the energy is coming from the particles themselves).
• It produces sharp spectral lines: emission spectrum (by hot low-density gas) and absorption spectrum (by cold low-density gas).

Answer: C

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

Nuclear & Particle Physics - EM Spectrum

Electromagnetic radiation provides a means to probe aspects of the physical universe. Which of the following statements regarding radiation spectra is NOT correct?

1. Lines in the infrared, visible, and ultraviolet regions of the spectrum reveal primarily the nuclear structure of the sample.
2. The wavelengths identified in an absorption spectrum of an element are among those in its emission spectrum.
3. Absorption spectra can be used to determine which elements are present in distant stars.
4. Spectral analysis can be used to identify the composition of galactic dust.
5. Band spectra are due to molecules.

(GR0177 #64)

Solution:

The EM spectrum of an object is the characteristic distribution of electromagnetic radiation emitted or absorbed by that particular object.

 EM spectrum can be used to determine the elements that compose the object, not the nuclear structure. (A) FALSE

Answer: A

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