The wavelength of spectral line coming from a distant star shifts from 600 nm to 600.1 nm. The velocity of the star relative to earth is

Options

A. 50 km `s^(-1)`
B. 100 km `s^(-1)`
C. 25 km `s^(-1)`
D. 200 km `s^(-1)`

Answer

A

Explanation

Here, `lambda = 600 nm`
`triangle lambda=600.1 nm - 600nm=0.1 nm`
As `(v_(s))/(c)= (triangle lambda)/(lambda) , :. (triangle lambda)/(lambda)c=(0.1nm)/(600nm)xx3xx10^(8)ms^(-1)`
` :.v_(s)=50xx10^(3)ms^(-1)=50kms^(-1)`