Answer:
the speed is 0.53 10⁸ m / s
Explanation:
As say in the exercise, the Doppler effect must be applied, in this case because it is an electromagnetic radiation with speed 3 10⁸ m/s and nothing can go faster we must use the relativistic Doppler effect
fo = fe √[(c-v) /(c + v)]
Where fo is the observed frequency, fe the emitted frequency, c the speed of light and v the relative speed of the observer and emit, it is positive move away
The light fulfills the relationship
c = λ f
f = c / λ
In this case v is negative since the source and the observer approach
Substituting in the equation
c /λo = c /λe √[ (c + v) / (c-v)]
λo = 510 10⁻⁹ m
λe = 610 10⁻⁹ m
We calculate the speed
(λe/λo)² = (c + v) / (c-v)
(λe /λo)² (c-v) = c + v
v +(λe /λo)² v = (λe /λo)² c -c
v [1 +(λe /λo)²] = c [(λe /λo)²-1]
v = c [(λe /λo)² -1] / [1 +(λe /λo)²]
v = 3 10⁸ [(610/510)² -1] / [1+ (610/510)²]
v = 3 10⁸ [1.43-1] / [1 + 1.43]
v = 0.53 10⁸ m / s
A little more about the speed of light
