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Step-by-Step Solution
Step 1: Identify the Known Values
• Speed of the galaxy, v = 286 \,\text{km s}^{-1} = 286 \times 10^{3} \,\text{m s}^{-1}
• Speed of light, c = 3 \times 10^{8} \,\text{m s}^{-1}
• Original wavelength of light (red line), \lambda = 630 \,\text{nm} = 630 \times 10^{-9} \,\text{m}
Step 2: Use the Doppler-Shift Formula for Light
For a source moving away at speed v (much smaller than c ), the shift in wavelength \Delta \lambda is given by:
\displaystyle \Delta \lambda = \lambda' - \lambda = \lambda \left(\frac{v}{c}\right)
Here, \lambda' is the observed wavelength and \lambda is the original wavelength.
Step 3: Substitute the Given Values
\displaystyle \Delta \lambda
= \left(630 \times 10^{-9}\,\text{m}\right)\Bigl(\frac{286 \times 10^{3}\,\text{m s}^{-1}}{3 \times 10^{8}\,\text{m s}^{-1}}\Bigr)
Step 4: Calculate the Numerical Value
First, multiply numerator terms:
630 \times 10^{-9} \times 286 \times 10^{3} = 630 \times 286 \times 10^{-9+3} = (630 \times 286) \times 10^{-6}
630 \times 286 = 180180 (you may approximate or do the exact multiplication).
So the numerator becomes 180180 \times 10^{-6} .
Now divide by 3 \times 10^{8} :
\displaystyle \frac{180180 \times 10^{-6}}{3 \times 10^{8}}
= 180180 \times 10^{-6} \times \frac{1}{3 \times 10^{8}}
= \frac{180180}{3} \times 10^{-6} \times 10^{-8}
= 60060 \times 10^{-14}
= 6.006 \times 10^{-10}\,\text{m} \approx 6 \times 10^{-10}\,\text{m}.
Step 5: Conclude the Value of x
The wavelength shift is of the form x \times 10^{-10}\,\text{m} . From the calculation:
\displaystyle x \approx 6.
Therefore, to the nearest integer, x = 6.
