An astronomer observes electromagnetic waves emitted by oxygen atoms in a distant galaxy that have a frequency of 5.710 × 10¹⁴ Hz. In the laboratory on earth, oxygen atoms emit waves that have a frequency of 5.841 × 10¹⁴ Hz. Determine the relative velocity of the galaxy with respect to the astronomer on the earth. Note: The speed of light is 2.9979 × 10⁸ m/s. A)6.724 × 10⁶ m/s, away from earth B)6.724 × 10⁶ m/s, toward earth C)2.931 × 10⁸ m/s, away from earth D)4.369 × 10⁴ m/s, toward earth E)4.369 × 10⁴ m/s, away from earth

Question

Quizplus · Accepted Answer

First, we need to calculate the difference in the frequency of the observed and laboratory waves:

Δf = f_observed - f_lab
Δf = 5.710 × 10^14 Hz - 5.841 × 10^14 Hz
Δf = -1.31 × 10^12 Hz

The negative sign indicates that the observed frequency is lower than the laboratory frequency, which means that the galaxy is moving away from the Earth.

Next, we can use the Doppler formula to find the relative velocity (v) of the galaxy:

Δf/f_lab = v/c
-1.31 × 10^12 Hz/5.841 × 10^14 Hz = v/2.9979 × 10^8 m/s
v = -6.724 × 10^16 m/s

The negative sign indicates that the galaxy is moving away from the Earth. Converting this to a more manageable unit of km/s, we get:

v = -6.724 × 10^16 m/s = -6.724 × 10^13 km/s = -67,240 km/s

Therefore, the answer is A) 6.724 × 10^16 m/s, away from earth.

An Astronomer Observes Electromagnetic Waves Emitted by Oxygen Atoms in a Distant