Question

An astronomer observes electromagnetic waves emitted by oxygen atoms in a distant galaxy that have a frequency of 5.710 × 10^14 Hz. In the laboratory on earth, oxygen atoms emit waves that have a frequency of 5.841 × 10^14 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.

Answer 1

`v = 6.72 * 10^6 m/s`

Step-by-step explanation:

As we know by the doppler's effect of light that when source moves with respect to stationary observer then the frequency of light received is different from its actual frequency

the relative change in the frequency of light is given by the formula

`(\Delta \\u)/(\\u) = (v)/(c)`

here we know that

`\Delta \\u = 5.841 * 10^(14) - 5.710* 10^14`

`\Delta \\u = 1.31 * 10^(13) Hz`

now we have

`(1.31 * 10^(13))/(5.841 * 10^(14)) = (v)/(2.9979 * 10^8)`

`v = 6.72 * 10^6 m/s`