The red shift of radiation froma distance nebula consits of light known to hace a wavelength of 434 nm. In the laboratory, this wavelength appears to be 6562 `Å.` What is the speed of the nebulal in the line of sight relative to the earth ? Is it approching or receding ?

Answer

`1.2 xx 10^(8)ms^(-1)` ; receding

Explanation

Here, `lambda = 434 nm = 434 xx 10^(-9) m`
`lambda' = 6562 Å = 6562 xx 10^(-10)m`.
As `v' = ((1 - v//c))/(sqrt(1 - v^(2)//c^(2))) = sqrt((1 - v//c)/(1 + v//c)) v`
`:. (1 - v//c)/(1 + v//c) = ((v')/(v))^(2) = ((lambda)/(lambda'))^(2) = ((434 xx 10^(-9))/(6562 xx 10^(-10)))^(2)`
`(1 - v//c)/(1 + v//c) = ((v')/(v))^(2) = 0.437`.
On solving, we get `v = 1.2 xx 10^(8)ms^(-1)`
As apparent wavelength increases, the nebula must be receding away from earth.