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. `2 xx10^(5)m//s`
B. `3xx10^(5)m//s`
C. `3xx10^(6)m//s`
D. `2.5 xx 10^(6)m//s`

Answer

C

Explanation

In this case,
`Delta lambda= 4747-4700=47`Å
`c=3xx10^(8)m//s lambda=4700`Å
According to Doppler's formula, the change in wavelength is given by
`Delta lambda =((v)/(c ))lambda`
where, v is the velocity of the star.
`therefore v=(c Delta lambda)/(lambda)=(3xx10^(8)xx47xx10^(-10))/(4700xx10^(-10))`
`therefore v = 3 xx 10^(6)m//s`