In hydrogen, the transition from level 2 to level 1 has a rest wavelength of 121.6 nm.

(A) Find the speed and direction for a star in which this line appears at wavelength 120.6 nm.

(B) Find the speed and direction for a star in which this line appears at wavelength 121.4 nm.

(C) Find the speed and direction for a star in which this line appears at wavelength 122.2 nm.

(D) Find the speed direction for a star in which this line appears at wavelength 123.0 nm.

A. v = 2,480 10⁶ m / s, The start away us

B. v = 4.96 105 m / s, The Star moves away

C. v = - 1.48 10⁶ m / s, The star is approaching

D. v = - 3.43 10⁶ m / s, The star is approaching

Explanation:

This variation in the wavelength of the light emitted by the Star is explained by the relativistic Doppler effect.

f' = √[ (1 + v / c) / (1-v / c) ] f

The speed of light is

c = λ f

f = c / λ

We replace

c / λ' = √ [ (1 + v / c) / (1-v / c) ] c / λ

Let's clear the speed, calling

β = v / c

1 / λ'² = [ (1 + β) / (1 - β) ] 1 / λ²

(1 - β) = (1 + β) λ'² / λ²

β (λ'² / λ² + 1) = 1 - λ'² / λ²

β = [ (1-λ'² / λ²) / (1 + λ'² / λ²) ]

v / c = [ (1 - λ'² / λ²) / (1 + λ'² / λ²) ]

v = c [ (1 - λ'² / λ²) / (1 + λ'² / λ²) ]

Now we can solve case house

To make the calculation easier, let's call a = λ' / λ

v = c [ (1 - a²) / (1 + a²) ]

A) λ = 121.6 nm

λ' = 120.6 nm

Let's replace

a² = (120.6 / 121.6) ²

a² = 0.9836

v = c [ (1 - 0.9836) / (1 + 0.9836) ]

v = 3 10⁸ 8.2678 10⁻³

v = 2,480 10⁶ m / s

The start away us

B) λ' = 121.4 nm

a² = (121.4 / 121.6) ²

a² = 0.9967

v = c [ (1 - 0.9967) / (1 + 0.9967) ]

v = 3 10⁸ 1.6527 10⁻³

v = 4.96 105 m / s

The Star moves away

C) λ' = 122.2 nm

a² = (122.2 / 121.6) ²

a² = 1.00989

v = c [ (1 - 1.00989) / (1 + 1.00989) ]

v = 3 10⁸ ( - 4.9207 10⁻³)

v = - 1.48 10⁶ m / s

The star is approaching

D) λ' = 123.0 nm

a² = (123.0 / 121.6) ²

a² = 1.02316

v = c [ (1 - 1.02316) / (1 + 1.02316) ]

v = 3 10⁸ (-1.1447 10⁻²)

v = - 3.43 10⁶ m / s

The star is approaching