One of the wavelengths of light emitted by hydrogen atoms under normal laboratory conditions is at ?0 = 656.3nm in the red portion of the electromagnetic spectrum. In the light emitted from a distant galaxy this same spectral line is observed to be Doppler-shifted to ? = 953.3nm , in the infrared portion of the spectrum.
How fast are the emitting atoms moving relative to the earth?


Sagot :

Answer:

1.07 × 10⁸ m/s

Explanation:

Using the relativistic Doppler shift formula which can be expressed as:

[tex]\lambda_o = \lambda_s \sqrt{\dfrac{c+v}{c-v}}[/tex]

here;

[tex]\lambda _o[/tex] = wavelength measured in relative motion with regard to the source at velocity v

[tex]\lambda_s =[/tex] observed wavelength from the source's frame.

Given that:

[tex]\lambda _o[/tex] = 656.3 nm

[tex]\lambda_s =[/tex] 953.3 nm

We will realize that [tex]\lambda _o[/tex] > [tex]\lambda_s[/tex]; thus, v < 0 for this to be true.

From the above equation, let's make (v/c) the subject of the formula: we have:

[tex]\dfrac{\lambda_o}{\lambda_s}=\sqrt{\dfrac{c+v}{c-v}}[/tex]

[tex]\Big(\dfrac{\lambda_o}{\lambda_s} \Big)^2=\dfrac{c+v}{c-v}[/tex]

[tex]\dfrac{v}{c} =\dfrac{\Big(\dfrac{\lambda_o}{\lambda_s} \Big)^2-1}{\Big(\dfrac{\lambda_o}{\lambda_s} \Big)^2+1}[/tex]

[tex]\dfrac{v}{c} =\dfrac{\Big(\dfrac{656.3}{953.3} \Big)^2-1}{\Big(\dfrac{656.3}{953.3} \Big)^2+1}[/tex]

[tex]\dfrac{v}{c} =0.357[/tex]

v = 0.357 c

To m/s:

1c = 299792458 m/s

0.357c = (299 792 458 × 0.357) m/s

= 107025907.5 m/s

= 1.07 × 10⁸ m/s