We know that the frequency of spectral line can be calculate as [f=frac{ c}{ lambda}] where $c= 3 cdot 10^{8} hspace{0.5mm} mathrm{frac{m}{s}}$ is the speed of light, and $lambda$ is the wavelength. In our case, the wavelength of oxygen’s spectral line is $ lambda=513 hspace{0.5mm} text{nm} Rightarrow lambda=513 cdot 10^{-9} hspace{0.5mm} text{m}$. So the frequency is
begin{align*}
f&=frac{c}{ lambda }\
f&=frac{3 cdot 10^{8} hspace{0.5mm} mathrm{frac{m}{s}}}{ 513 cdot 10^{-9} hspace{0.5mm} mathrm{m}}\
f&= frac{3}{513} cdot 10^{17} hspace{0.5mm} frac{1}{mathrm{s}}\
f&= 584.8 cdot 10^{12} hspace{0.5mm} text{Hz}
end{align*}
So, the frequency of oxygen’s spectral line is [ framebox[1.1width]{$ therefore f= 584.8 cdot 10^{12} hspace{0.5mm} text{Hz} $}]

$$
f= 584.8 cdot 10^{12} hspace{0.5mm} text{Hz}
$$

The speed of a hydrogen atom in a galaxy is $6.55 cdot 10^{6} hspace{0.5mm} mathrm{frac{m}{s}}$. That atom emits light with frequency of $6.16 cdot 10^{14} hspace{0.5mm} text{Hz}$, while is moving away from Earth. To find a frequency of light that would be observed by astronomer on Earth, we have to use equation $f_{obs}=f hspace{0.5mm} (1 – frac{v}{c} )$, where $v$ is the speed of a hydrogen atom in a galaxy, $c$ is the speed of the light, and $f$ is frequency of light that atom emits.
begin{align*}
f_{obs}&=f hspace{0.5mm} left(1 – frac{v}{c}right )\
f_{obs}&=6.16 cdot 10^{14} hspace{0.5mm} text{Hz} hspace{0.5mm} left(1 – frac{6.55 cdot 10^{6} hspace{0.5mm} mathrm{frac{m}{s}}}{3.00 cdot 10^{8} hspace{0.5mm} mathrm{frac{m}{s}}}right )\
f_{obs}&=6.16 cdot 10^{14} hspace{0.5mm} text{Hz} hspace{0.5mm} (1 – 2.18 cdot 10^{-2})\
f_{obs}&=6.15 cdot 10^{14} hspace{0.5mm} text{Hz}\
end{align*}
The frequency of light that would be observed by astronomer on Earth is [ framebox[1.1width]{$ therefore f=6.15 cdot 10^{14} hspace{0.5mm} text{Hz} $}]

$$
f=6.15 cdot 10^{14} hspace{0.5mm} text{Hz}
$$

The speed of a hydrogen atom in a galaxy is $6.55 cdot 10^{6} hspace{0.5mm} mathrm{frac{m}{s}}$ (cannot be $6.55 cdot 10^{16} hspace{0.5mm} mathrm{frac{m}{s}}$ because it is greater than the speed of light). That atom emits light with wavelength of $4.86 cdot 10^{-7} hspace{0.5mm} mathrm{m}$, while is moving away from Earth. To calculate the wavelength that would be observed on Earth, we start from the equation $f_{obs}=f hspace{0.5mm} (1 – frac{v}{c} )$, and use the equation $f= frac{ c}{ lambda}$, where $c$ is the speed of light, and $ lambda$ is the wavelength.
begin{align*}
f_{obs}&=f hspace{0.5mm} (1 – frac{v}{c} )\
frac{c}{ lambda_{obs}}&= frac{c}{ lambda} hspace{0.5mm} (1 – frac{v}{c} ) /:c\
lambda &= lambda_{obs} hspace{0.5mm} (1 – frac{v}{c} )\
lambda_{obs} &= lambda hspace{0.5mm} (1 – frac{v}{c} )^{-1}\
lambda_{obs} &= 4.86 cdot 10^{-7} hspace{0.5mm} mathrm{m} hspace{0.5mm} (1 – frac{6.55 cdot 10^{6} hspace{0.5mm} mathrm{frac{m}{s}}}{3.00 cdot 10^{8} hspace{0.5mm} mathrm{frac{m}{s}}} )^{-1}\
lambda_{obs} &= 4.86 cdot 10^{-7} hspace{0.5mm} mathrm{m} hspace{0.5mm} (1 – 2.18 cdot 10^{-2} )^{-1}\
lambda_{obs} &= 4.97 cdot 10^{-7} hspace{0.5mm} mathrm{m}
end{align*}
The wavelength of light that would be observed by astronomer on Earth is [ framebox[1.1width]{$ therefore lambda_{obs} = 4.97 cdot 10^{-7} hspace{0.5mm} mathrm{m} $}]

$$
lambda_{obs} = 4.97 cdot 10^{-7} hspace{0.5mm} mathrm{m}
$$

The wavelength of oxygen’s spectral line is $ lambda= 513 hspace{0.5mm} mathrm{nm}$, and the observed wavelength of oxygen’s spectral line is $ lambda_{obs}=525 hspace{0.5mm} mathrm{nm} $. To find whether the galaxy is moving toward or away from Earth, we observe the equation for Doppler Shift $( lambda_{obs}- lambda= pm frac{v}{c} lambda)$. The part $ frac{v}{c}$ should be greater than zero. We can write
begin{align*}
lambda_{obs}- lambda&= pm frac{v}{c} lambda / : lambda\
frac{ lambda_{obs}}{ lambda} -1 &= pm frac{v}{c}\
frac{ 525 hspace{0.5mm} mathrm{nm} }{ 513 hspace{0.5mm} mathrm{nm}} -1 &= pm frac{v}{c}\
1.0234-1&= pm frac{v}{c}\
+ frac{v}{c}&=0.0234\
v&=0.0234c\
v&=0.0234 cdot 3 cdot 10^8 hspace{0.5mm} mathrm{frac{m}{s}}\
v&=7 cdot 10^6 hspace{0.5mm} mathrm{frac{m}{s}}
end{align*}
The sign was positive, so the galaxy is moving away from Earth, and the speed of oxygen atom is [ framebox[1.1width]{$ therefore v=7 cdot 10^6 hspace{0.5mm} mathrm{frac{m}{s}} $}]

$$
v=7 cdot 10^6 hspace{0.5mm} mathrm{frac{m}{s}}
$$