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 - 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