Question

If you found a galaxy with an Ha emission line that had a wavelength of 756.3 nm, what would be the galaxy’s distance if the Hubble constant is 70 km/s/Mpc? (Note that the rest wavelength of the Ha emission line is 656.3 nm.)

Answer 1

Answer: 653.011Mpc

Explanation:

Hubble deduced that the farther the galaxy is, the more redshifted it is in its spectrum, and noted that all galaxies are "moving away from each other with a speed that increases with distance", and enunciated the now called Hubble–Lemaître Law.  

This is mathematically expressed as:

$V=H_{o}D$ (1)

Where:

$V$ is the approximate recession velocity of the galaxy

$H_{o}=70 km/s/Mpc$ is the current Hubble constant  

$D$ is the galaxy's distance

On the other hand, the equation for the Doppler shift is:

$\frac{\Delta \lambda}{\lambda_{o}}=\frac{V}{c}$  (2)

Where:

$\lambda_{o}=656.3nm=656.3(10)^{-9}m$ is the wavelength for the Ha line of the galaxy observed at rest

$\Delta \lambda=\lambda_{1}-\lambda_{o}$ is the variation between the measured wavelength for the Ha emission line in the spectrum of this galaxy ($\lambda_{1}=756.3nm=756.3(10)^{-9}m$ in this case) and the wavelength for the same Ha line observed at rest

$c=3(10)^{8}m/s$ is the speed of light

Rewriting (2):

$\frac{\lambda_{1}-\lambda_{o}}{\lambda_{o}}=\frac{V}{c}$  (3)

Isolating $V$:

$V=\frac{(\lambda_{1}-\lambda_{o})c}{\lambda_{o}}$  (4)

Finding $V$:

$V=\frac{(756.3(10)^{-9}m-656.3(10)^{-9}m)3(10)^{8}m/s}{656.3(10)^{-9}m}$  (5)

$V=45710802.99m/s=45710.80299km/s$  (6)

Substituting $V$ in (1):

$45710.80299km/s=(70km/s/Mpc)D$ (7)

Finding $D$:

$D=\frac{V}{H_{o}}=\frac{45710.80299km/s}{70km/s/Mpc}$ (8)

Finally:

$D=653.011Mpc$ (8)  This is the galaxy's distance