In the spectrum of light of a luminous heavenly body the wavelength of a spectral line is measured to be `4747 Å` while actual wavelength of the line is `4700 Å`. The relative velocity of the heavenly body with respect to earth will be (velocity of light is `3 xx 10^(8) m//s`)

Options

A. `2xx10^(5) m//s `
B. `3xx10^(5) m//s `
C. `3xx10^(6) m//s `
D. `2.5xx10^(6) m//s `

Answer

C

Explanation

In this case `, Delta lambda=4747-4700=47Å`
`c=3 xx 10^(8) m//s, lambda -4700 Å`
According to Doppler's formula, the change in wavelength is given by
`Deltalambda=((v)/(c))lambda ` where v is the velocity of the star.
`:. v=(c Delta lambda)/(lambda)=(3 xx 10^(8) xx 47 xx 10^(-10))/( 4700 xx 10^(-10))`
`:. v=3 xx 10^(6) m//s `