Question

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?

Answer 1

1.07 × 10⁸ m/s

Step-by-step explanation:

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

`\lambda_o = \lambda_s \sqrt{(c+v)/(c-v)}`

here;

`\lambda _o` = wavelength measured in relative motion with regard to the source at velocity v

`\lambda_s =` observed wavelength from the source's frame.

Given that:

`\lambda _o` = 656.3 nm

`\lambda_s =` 953.3 nm

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

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

`(\lambda_o)/(\lambda_s)=\sqrt{(c+v)/(c-v)}`

`\Big((\lambda_o)/(\lambda_s) \Big)^2=(c+v)/(c-v)`

`(v)/(c) =(\Big((\lambda_o)/(\lambda_s) \Big)^2-1)/(\Big((\lambda_o)/(\lambda_s) \Big)^2+1)`

`(v)/(c) =(\Big((656.3)/(953.3) \Big)^2-1)/(\Big((656.3)/(953.3) \Big)^2+1)`

`(v)/(c) =0.357`

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