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

Step-by-step 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:

`(\Delta \lambda)/(\lambda_(o))=(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):

`(\lambda_(1)-\lambda_(o))/(\lambda_(o))=(V)/(c)` (3)

Isolating
`V`:

`V=((\lambda_(1)-\lambda_(o))c)/(\lambda_(o))` (4)

Finding
`V`:

`V=((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=(V)/(H_(o))=(45710.80299km/s)/(70km/s/Mpc)` (8)

Finally:

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