Answer
The distance between the galaxy B and the Earth when the light was emitted is $~~1.0\times 10^9~ly$
Work Step by Step
In part (b), we found that $\frac{\Delta \lambda}{\lambda} = \frac{r~\alpha}{c-r~\alpha}$
We can find the distance between galaxy B and the Earth when the light was emitted:
$\frac{\Delta \lambda}{\lambda} = \frac{r~\alpha}{c-r~\alpha} = 0.080$
$r~\alpha = (0.080)(c-r~\alpha)$
$r~\alpha+0.080~r~\alpha = 0.080~c$
$r = \frac{0.080~c}{\alpha(1+0.080)}$
$r = \frac{0.080~c}{1.080~H}$
$r = \frac{(0.080)(3.0\times 10^8~m/s)}{(1.080)~(0.0218~m/s~ly)}$
$r = 1.0\times 10^9~ly$
The distance between the galaxy B and the Earth when the light was emitted is $~~1.0\times 10^9~ly$
