What is the observed wavelength of the 656.3 nm (first Balmer) line of hydrogen emitted by a galaxy at a distance of 2.40 x 108 ly

λ = 667.85 nm

Explanation:

Let f be the frequency detected by the observer

Let v be the speed at which the observer is moving.

Now, when the direction at which the observer is moving is away from the source, we have the frequency as;

f = f_o√((1 - β)/(1 + β))

From wave equations, we know that the wavelength is inversely proportional to the frequency. Thus, wavelength is now;

λ = λ_o√((1 + β)/(1 - β))

Where, β = Hr/c

H is hubbles constant which has a value of 0.0218 m/s • ly

c is speed of light = 3 × 10^(8) m/s

r is given as 2.40 x 10^(8) ly

Thus,

β = (0.0218 × 2.4 x 10^(8))/(3 × 10^(8))

β = 0.01744

Since we are given λ_o = 656.3 nm

Then;

λ = 656.3√((1 + 0.01744)/(1 - 0.01744))

λ = 667.85 nm

1 - wave length

2 - amplitude

3 - crest

4 - trough

explanation:

duh.