The Doppler shift formula for the wavelength change when the source is moving away from the observer is: \[ \frac{\Delta \lambda}{\lambda} = \frac{v}{c} \] where: - \( \Delta \lambda = \lambda' - \lambda \) is the change in wavelength, - \( v \) is the velocity of the galaxy relative to Earth, - \( c \) is the speed of light. Substituting the values: \[ \frac{601 - 600}{600} = \frac{v}{3 \times 10^8} \] \[ \frac{1}{600} = \frac{v}{3 \times 10^8} \] Solving for \( v \): \[ v = \frac{3 \times 10^8}{600} = 500 \, \text{km/s} \]
The speed of the galaxy with respect to Earth is \( {500 \, \text{km/s}} \), so the correct answer is (A).
This problem involves the use of the Doppler effect for light. The formula for the change in wavelength due to the relative motion of a source and observer is given by: \[ \frac{\Delta \lambda}{\lambda_0} = \frac{v}{c} \] Where:
\(\Delta \lambda\) = \(\lambda_{\text{observed}}\)
\(\lambda_{\text{rest}}\) is the change in wavelength,
\(\lambda_0\) is the rest wavelength,
\(v\) is the velocity of the source (the galaxy),
\(c\) is the speed of light in vacuum.
From the question: \[ \lambda_0 = 600 \, \text{nm}, \quad \lambda_{\text{observed}} = 601 \, \text{nm} \] Thus: \[ \Delta \lambda = 601 \, \text{nm} - 600 \, \text{nm} = 1 \, \text{nm} \] Substituting the values into the Doppler formula: \[ \frac{1 \, \text{nm}}{600 \, \text{nm}} = \frac{v}{3 \times 10^8 \, \text{m/s}} \] Solving for \(v\): \[ v = \frac{1}{600} \times 3 \times 10^8 \, \text{m/s} = 5 \times 10^5 \, \text{m/s} = 500 \, \text{km/s} \] Thus, the speed of the galaxy with respect to the Earth is \(500 \, \text{km/s}\).