A distant star is receding from the Earth with a speed of 1.4o x 107 m/s. It emits light of frequency 4.57 x 10^14 Hz. The speed of light is 3.0 x 10^8 m/s. The Doppler effect formula can be used with light waves. What will bethe frequency of this light when detected on Earth?
Correct answer: B. 4.36 x 10^14 Hz
- A. 2.04 x 10^13 Hz
- B. 4.36 x 10^14 Hz
- C. 4.57 x 10^14 Hz
- D. 4.79 x 10^14 Hz
Explanation
Given: v (receding speed) = 1.40 x 10^7 m/s f_source (emitted frequency) = 4.57 x 10^14 Hz c (speed of light) = 3.0 x 10^8 m/s Doppler effect formula for light: f_observed = f_source * sqrt((1 - v/c) / (1 + v/c)) Calculations: v/c = (1.40 x 10^7 m/s) / (3.0 x 10^8 m/s) = 0.0466666... sqrt((1 - v/c) / (1 + v/c)) = sqrt((1 - 0.0466666...) / (1 + 0.0466666...)) sqrt((1 - v/c) / (1 + v/c)) = sqrt(0.9533333 / 1.0466666) sqrt((1 - v/c) / (1 + v/c)) = sqrt(0.9108108) sqrt((1 - v/c) / (1 + v/c)) = 0.9543641 (approximately) f_observed = 4.57 x 10^14 Hz * 0.9543641 f_observed = 4.361936 x 10^14 Hz Result: f_observed = 4.36 x 10^14 Hz (approximately)
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