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.
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