Answer:
D. The star is approaching Earth.
Explanation:
As we know by the doppler's effect of light that
[tex]\frac{\Delta \nu}{\nu} = \frac{v}{c}[/tex]
here we know that
[tex]\Delta \nu[/tex] = change in frequency
here we know that the wavelength of light coming from the star is decreased so the frequency will increase
[tex]\Delta \nu = \frac{c}{\lambda'} - \frac{c}{\lambda}[/tex]
[tex]\Delta \nu =(3\times 10^8)(\frac{1}{6.56186 \times 10^{-7}} - \frac{1}{6.563 \times 10^{-7}})[/tex]
[tex]\Delta \nu = 7.9414 \times 10^{10} Hz[/tex]
now we have
[tex]\frac{7.9414 \times 10^{10}}{\nu} = \frac{v}{c}[/tex]
here we know that
[tex]\nu = \frac{3\times 10^8}{6.563 \times 10^{-7}} = 4.57 \times 10^{14} Hz[/tex]
now we have
[tex]v = 5.2 \times 10^4 m/s[/tex]
So here correct answer is
D. The star is approaching Earth.
