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The red shift of radiation from a distant nebula consists of light known to have a wavelength of 434 nm. In the laboratory, this wavelength appears to be 6562 Å. What is the speed of the nebula in the line of sight relative to the earth ? Is it approaching or receding ? |
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Answer» `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. |
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