Question:

If a molecule emitting a radiation of frequency \(3.100 \times 10^9\) Hz approaches an observer with a relative speed of \(5.000 \times 10^6\) m s\(^{-1}\), then the observer detects a frequency of ___________ \( \times 10^9 \) Hz. (rounded off to three decimal places) [Given: Speed of light \(c = 3.000 \times 10^8\) m s\(^{-1}\)]

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For a source approaching an observer, the observed frequency is higher than the emitted frequency (blueshift). For a source moving away, the observed frequency is lower (redshift). The relativistic Doppler effect formula should be used when the speeds are a significant fraction of the speed of light, but it is also generally applicable at lower speeds.
Updated On: Apr 19, 2025
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Solution and Explanation

This problem involves the Doppler effect for electromagnetic radiation. When a source of radiation moves relative to an observer, the observed frequency (\(f'\)) is different from the emitted frequency (\(f\)). For a source approaching an observer, the relativistic Doppler effect formula is: \[ f' = f \sqrt{\frac{1 + v/c}{1 - v/c}} \] where \(f\) is the emitted frequency, \(v\) is the relative speed of the source and the observer, and \(c\) is the speed of light. Given values: Emitted frequency \(f = 3.100 \times 10^9\) Hz Relative speed \(v = 5.000 \times 10^6\) m s\(^{-1}\) Speed of light \(c = 3.000 \times 10^8\) m s\(^{-1}\) First, calculate the ratio \(v/c\): \[ \frac{v}{c} = \frac{5.000 \times 10^6}{3.000 \times 10^8} = \frac{5}{300} = \frac{1}{60} \approx 0.01667 \] Now, substitute this into the Doppler effect formula: \[ f' = (3.100 \times 10^9) \sqrt{\frac{1 + 1/60}{1 - 1/60}} = (3.100 \times 10^9) \sqrt{\frac{61/60}{59/60}} \] \[ f' = (3.100 \times 10^9) \sqrt{\frac{61}{59}} = (3.100 \times 10^9) \sqrt{1.033898} \] \[ f' = (3.100 \times 10^9) \times 1.016807 \] \[ f' = 3.1521017 \times 10^9 \, \text{Hz} \] Rounding off to three decimal places for the factor multiplying \(10^9\), we get 3.152. So, the observed frequency is \(3.152 \times 10^9\) Hz. This value (3.152) falls within the given correct answer range of 3.110 to 3.200. \bigskip
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