a hydrogen line in a stars spectrum has a frequency of 4.57×10^14 hz when stationary. in altairs spectrum…

Explanation:

Step1: Identify the Doppler - shift formula for light

For light, the Doppler - shift formula when the source is moving away or towards the observer is $\Delta f=\frac{v}{c}f_0$, where $\Delta f$ is the change in frequency, $v$ is the velocity of the source relative to the observer, $c = 3\times10^{8}\ m/s$ is the speed of light in vacuum, and $f_0$ is the rest - frequency.

Step2: Rearrange the formula to solve for $v$

From $\Delta f=\frac{v}{c}f_0$, we can solve for $v$ as $v = c\times\frac{\Delta f}{f_0}$.

Step3: Substitute the given values

We are given that $f_0=4.57\times 10^{14}\ Hz$ and $\Delta f = 3.98\times 10^{10}\ Hz$, and $c = 3\times10^{8}\ m/s$. Substituting these values into the formula $v = c\times\frac{\Delta f}{f_0}$, we get $v=(3\times 10^{8})\times\frac{3.98\times 10^{10}}{4.57\times 10^{14}}$. First, calculate $\frac{3.98\times 10^{10}}{4.57\times 10^{14}}=\frac{3.98}{4.57}\times10^{10 - 14}\approx0.871\times10^{- 4}$. Then, $v=(3\times 10^{8})\times(0.871\times10^{-4})$. Using the rule of exponents $a^m\times a^n=a^{m + n}$, we have $v=(3\times0.871)\times10^{8+( - 4)}=2.613\times10^{4}\ m/s$.

Answer:

$2.61\times 10^{4}\ m/s$