This problem involves calculating the observed frequency of radiation emitted by a source approaching an observer, which is an application of the Doppler effect.

Doppler Effect Calculation for Approaching Source

When a source of radiation moves towards an observer, the observed frequency ($f_{obs}$) is higher than the emitted frequency ($f_{source}$). For speeds significantly less than the speed of light ($v \ll c$), the non-relativistic Doppler effect formula can be used.

Applying the Doppler Effect Formula

The formula for the observed frequency when the source approaches the observer is:

$f_{obs} = f_{source} \left(1 + \frac{v}{c}\right)$

Where:

• $f_{obs}$ is the observed frequency.
• $f_{source}$ is the frequency emitted by the source.
• $v$ is the relative speed between the source and the observer.
• $c$ is the speed of light.

Step-by-Step Calculation

Given values:

• Source frequency, $f_{source} = 3.100 \times 10^9 \text{ Hz}$
• Relative speed, $v = 5.000 \times 10^6 \text{ m s}^{-1}$
• Speed of light, $c = 3.000 \times 10^8 \text{ m s}^{-1}$

First, calculate the ratio $\frac{v}{c}$:

$\frac{v}{c} = \frac{5.000 \times 10^6 \text{ m s}^{-1}}{3.000 \times 10^8 \text{ m s}^{-1}} = \frac{5}{300} = \frac{1}{60}$

Now, substitute this value into the Doppler effect formula:

$f_{obs} = (3.100 \times 10^9 \text{ Hz}) \left(1 + \frac{1}{60}\right)$

$f_{obs} = (3.100 \times 10^9 \text{ Hz}) \left(\frac{60+1}{60}\right)$

$f_{obs} = (3.100 \times 10^9 \text{ Hz}) \left(\frac{61}{60}\right)$

Calculate the final observed frequency:

$f_{obs} = 3.100 \times \frac{61}{60} \times 10^9 \text{ Hz}$

$f_{obs} = \frac{189.1}{60} \times 10^9 \text{ Hz}$

$f_{obs} = 3.151666... \times 10^9 \text{ Hz}$

Rounding the coefficient to three decimal places:

$f_{obs} \approx 3.152 \times 10^9 \text{ Hz}$

The observer detects a frequency of 3.152 $\times 10^9 \text{ Hz}$.