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
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