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Step-by-Step Solution
Step 1: Identify the Concept
The observed shift in the wavelength of light from a receding galaxy can be explained by the Doppler effect for electromagnetic waves. When the galaxy moves away, the wavelength of the spectral line increases (red shift).
Step 2: Write Down the Relevant Doppler Formula (Approximate for Small Speeds)
For speeds much less than the speed of light (v ≪ c), the Doppler shift in wavelength can be approximated by:
$ \frac{\Delta \lambda}{\lambda_0} = \frac{v}{c} $
where
$\lambda_0$ is the original wavelength (at rest)
$\Delta \lambda$ is the change in wavelength ($\lambda - \lambda_0$)
$v$ is the speed of the source relative to the observer
$c$ is the speed of light in vacuum
Step 3: Substitute the Known Values
Given:
Original wavelength, $ \lambda_0 = 5890 \, \mathop{Å} $
Observed wavelength, $\lambda = 5896 \, \mathop{Å} $
$ \Delta \lambda = \lambda - \lambda_0 = 5896 - 5890 = 6 \, \mathop{Å} $
Speed of light, $ c \approx 3 \times 10^5 \text{ km/s} $ (using km/s directly to get final speed in km/s)
Step 4: Calculate the Speed of Recession
Using the approximate Doppler relation:
$ \frac{\Delta \lambda}{\lambda_0} = \frac{v}{c} \quad \Longrightarrow \quad v = c \times \frac{\Delta \lambda}{\lambda_0}
$
Substitute the numerical values:
$ v = \bigl(3 \times 10^5 \text{ km/s}\bigr) \times \frac{6}{5890}
$
Perform the division:
$ \frac{6}{5890} \approx 1.02 \times 10^{-3}
$
Hence,
$ v \approx 3 \times 10^5 \times 1.02 \times 10^{-3} \approx 306 \text{ km/s}
$
Step 5: State the Final Answer
The galaxy should move outward at approximately 306 km/s for the sodium-D line to shift from 5890 Å to 5896 Å as observed from Earth.
