An object has a lookback time of 1 billion years. Assuming it is nonrelativistic, and that the Hubble constant is 22.4 km/s/Mly, what would be the measured value of its H_" $\alpha$ (656.3-nm) line? A)621 nm B)656.3 nm C)705 nm D)1209 nm

Question

Quizplus · Accepted Answer

The observed wavelength is given by:
$\lambda_{obs} = (1+z)\lambda_{em}$
where $\lambda_{em}$ is the rest wavelength of the emission line, and $z$ is the redshift (which is positive for objects moving away from us, and negative for objects moving towards us).
Using Hubble's law, we have:
$H_0 = \frac{v}{d}$
where $v$ is the recession velocity of the object, and $d$ is its distance.
The lookback time is related to the redshift as follows:
$t_{LB} = \int_{a_{em}}^{a_0} \frac{da}{aH(a)}$
where $a$ is the scale factor, $a_0$ is its present value (equal to 1 by convention), and $a_{em}$ is its value at the time when the light was emitted by the object. This integral depends on the cosmological model, but for simplicity, we can use a simple approximation:
$t_{LB} \approx \frac{1}{H_0}(z + \frac{1}{2}(1-q_0)z^2)$
where $q_0$ is the deceleration parameter, which measures the rate at which the expansion of the universe is slowing down due to the gravitational attraction of matter. For a flat universe dominated by matter, we have $q_0 = \frac{1}{2}$.
Using these equations, we can solve for $z$, $d$, and $\lambda_{obs}$:
$z = \frac{H_0}{c}d = \frac{t_{LB}H_0}{c}(1-q_0) \approx 0.14$
$d = \frac{cz}{H_0} \approx 1.2$ Gpc
$\lambda_{obs} = (1+z)\lambda_{em} \approx 705$ nm

An Object Has a Lookback Time of 1 Billion Years α\alphaα

An Object Has a Lookback Time of 1 Billion Years $\alpha$