If the spectrum of a distant galaxy is observed to have a calcium K absorption line that occurs at a wavelength of 500.4 nm,what is this galaxy's distance if the rest wavelength of this absorption line is 393.4 nm? Assume a Hubble constant of 72 km/s/Mpc.
A) 950 Mpc
B) 1,800 Mpc
C) 2,300 Mpc
D) 1,130 Mpc

Question

Quizplus · Accepted Answer

The redshift (z) can be calculated using the formula $z = \frac{\lambda_{observed} - \lambda_{rest}}{\lambda_{rest}}$, where $\lambda_{observed}$ is the observed wavelength and $\lambda_{rest}$ is the rest wavelength. Substituting the given values, $z = \frac{500.4 - 393.4}{393.4} \approx 0.272$. The distance (D) can then be found using Hubble's Law, $v = H_0 \times D$, where $v$ is the velocity (calculated as $v = z \times c$, with $c$ being the speed of light, approximately $3 \times 10^5$ km/s), and $H_0$ is the Hubble constant. Rearranging for $D$, we get $D = \frac{v}{H_0} = \frac{z \times c}{H_0}$. Substituting the values, $D = \frac{0.272 \times 3 \times 10^5}{72}$ gives a distance of approximately 1130 Mpc.

If the Spectrum of a Distant Galaxy Is Observed to Have