The galaxy NGC 4414 is moving away from us at 7.16 × 10^5 m/s. By how much is the frequency of the 6.91 × 10^14 Hz hydrogen line shifted in the galaxy’s spectrum?
For a receding source, the Doppler frequency shift is approximately $\Delta f/f \approx -v/c$. With $v = 7.16\times 10^5\ \text{m/s}$ and $f = 6.91\times 10^{14}\ \text{Hz}$, $$\Delta f \approx -\left(\frac{7.16\times 10^5}{3.00\times 10^8}\right)(6.91\times 10^{14}) \approx -1.65\times 10^{12}\ \text{Hz}.$$ The line is shifted lower in frequency by $1.65\times 10^{12}\ \text{Hz}$.
What this question is asking
NGC 4414 is moving away from us, so its spectral lines are redshifted. In frequency terms, redshift means the observed frequency is smaller than the emitted (rest) frequency.
Using the Doppler shift for light (small-speed approximation)
Since $v \ll c$, we can use the linear Doppler approximation: $$\frac{\Delta f}{f} \approx -\frac{v}{c}$$ The minus sign is there because moving away decreases frequency.
Plug in the numbers
Compute the ratio: $$\frac{v}{c} = \frac{7.16\times 10^5}{3.00\times 10^8} \approx 2.39\times 10^{-3}$$ Now multiply by $f$: $$\Delta f \approx -\left(2.39\times 10^{-3}\right)\left(6.91\times 10^{14}\right) \approx -1.65\times 10^{12}\ \text{Hz}$$
Direction check (does the sign make sense?)
Because the galaxy is receding, the observed frequency must go down. So $\Delta f$ should be negative, which matches the calculation.
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