# Radiotelescopes and Wavelengths

Radio telescopes are designed to observe 21-cm waves emitted by atomic hydrogen gas. A signal from a distant radio-emitting galaxy is found to have a wavelength that is 0.28cm longer than the normal 21-cm wavelength.

Estimate the distance to this galaxy.

Express your answer using two significant figures.

d= ??? (units are Mly)

Answer

the redshift is

$z +1= \lambda(obs)/\lambda_0 =21.28/21 =1.01333$

$z =0.01333$

the speed is

$1+z = \sqrt{(1+v/c)/(1-v/c)}$

$1.02684 =(1+x)/)(1-x)$

$1.02684 -1.02684*x =1+x$

$0.02684 =2.02684*x$

$x =0.01324$

$v =0.01324*c =3972.7 km/s$

Now the relation between speed and distance is

$V = H*d$

where H is the Hubble constant $H =67.80 km/(s*Mpc)$

Mpc is Megaparsec

(http://en.wikipedia.org/wiki/Hubble’s_law)

The distance d is

$d = V/H = 3972.7/67.80 =58.594 Mpc =58.594*3.27 MLy =191.11 MLy$

(MLy=Mega Light years)

(http://www.metric-conversions.org/length/parsecs-to-lightyears.htm)