Answer:
Solved
Explanation:
due to Doppler effect on light
[tex]v= (1-\frac{\lambda}{\lambda_0})c[/tex]
v- radiation velocity of approach
c= velocity of light
λ=522.5 nm wavelength from the distant star
λ_0 = 486.1 rest wavelength
putting these values to find v
[tex]v= (1-\frac{522.5}{486.1})c[/tex]
=0.0748817*3*10^8
=1.92×10^8 m/ses
=1.92×10^5 Km/sec
since the sign of velocity is negative it is moving away from the Earth.
The radial velocity will be "[tex]1.92\times 10^5[/tex] km/sec".
It defines the frequency variations of any type of sound or light wave produced by a moving source throughout the relation to something like an observer.
Waves released by such an item moving towards another observer are compressed, resulting in a greater frequency as even the source gets closer towards the viewer.
According to the question,
Wavelength from distant star,
[tex]\lambda = 522.5[/tex] nm
Rest wavelength,
[tex]\lambda_0 = 486.1[/tex] nm
Velocity of light,
c = [tex]3\times 10^8[/tex]
By using the Doppler effect,
→ v = [tex](1- \frac{\lambda}{\lambda_0} )c[/tex]
By substituting the values,
= [tex](1-\frac{522.5}{486.1} )\times 3\times 10^8[/tex]
= [tex]0.0748817\times 3\times 10^8[/tex]
= [tex]1.92\times 10^8[/tex] m/s
or,
= [tex]1.92\times 10^5[/tex] km/s
Thus the above answer is right.
Find out more information about Doppler effect here:
https://brainly.com/question/4052291
