The hydrogen atoms in a star are also moving at high velocity because of the random motions caused by their high temperature. as a result, each atom is doppler shifted a little bit differently, leading to a finite width of each spectral line, such as the 656.46-nm line we were just discussing. for a star like our sun, this leads to a finite width of the spectral lines of roughly δλ=0.04nm. if our instruments can only resolve to this accuracy, what is the lowest speed v, greater than 0, that we can measure a star to be moving?

the lowest speed that can be measured is  ±  1,828 10⁴ m/s

Explanation:

The expression for the relativistic doppler effect is

        f₀ = fs √ (1-v / c) / (1 + v / c)

Where f₀ and fs are the observed and emitted frequency respectively, v is the velocity of the wave or particle

Let's use the relationship of the speed of the wave with its frequency and wavelength

    c = λ f

    f = c / λ

Let's replace

   c / λ₀ = c / λs √ (1-v / c) / (1 + v / c)

   λs/λ₀ = √ (1-v / c) / (1 + v / c)

We cleared the speed

    (λs / λ₀)² = (1-v / c) / (1 + v / c)

    (λs/λ₀)²2 = (c-v) / c + v)

    (c + v) (λs / λ₀)² = c-v

    v [1 + (λs / λ₀)²] = c [1 - (λs /λ₀)²]

    v = c [1 - (λs /λ₀)²] / [1 + (λs / λ₀)²]

    v = c (λ₀² - λs²) / (λ₀² +λs²)

Let's calculate the speeds for the two possible cases

When the wavelength increases

    λs = 656.46 + 0.04 = 656.50 nm

    λ₀ = 656.46 nm

   (λ₀² - λs²) = 656.5² - 656.46² = 52.518

   (λ₀² + λs²) = 656.5² + 656.46² = 861931.98

   v = 3 10⁸ 52.518 / 861931.98

   v = 1.828 10⁴ m / s

Case 2 the wavelength decreases

     λ₀ = 656.46 - 0.04 = 656.42 nm

   (λ₀² - λs²) = 656.42² - 656.46² = -52.5151

   (λ₀² + λs²) = 656.42² + 656.46² = 861826.948

   v = 3 10⁸ (- 52.5151) / 861826.948

   v = - 1,828 10⁴ m / s

The negative sign indicates that the Source moves away from the observer

b.transmitters send radio waves, and receivers capture radio waves.

spectral lines lyman is one set