Answer
$4.7\times10^9$ light-years.
Work Step by Step
Use equation 33–3 to express the ratio of wavelengths.
$$\frac{\lambda_{obs}}{\lambda_{rest}}=\sqrt{\frac{1+v/c}{1-v/c}}=\frac{610nm}{434nm}=1.40553$$
$$\sqrt{\frac{1+v/c}{1-v/c}}=1.40553$$
$$ \frac{1+v/c}{1-v/c}=1.9755$$
$$1.9755(1-v/c)=1+v/c$$
$$1.9755-1.9755v/c=1+v/c$$
$$2.9755v/c=0.9755$$
$$v=0.3278c$$
Find the distance from Hubble’s law, equation 33–4.
$$d=\frac{v}{H_0}$$
$$d=\frac{0.3278(3.00\times10^8m/s)}{ (21000m/s/Mly)}=4683\;Mly\approx4.7\times10^9\; ly$$
