n hydrogen, the transition from level 1 to level 2 has a rest wavelength of 121.6 nm. Suppose you see this line at a wavelength of 124.2 nm in a Star. What is the velocity of this in the direction of the Earth?

Respuesta :

Answer:

velocity of this in the direction of the Earth is 6.4 × [tex]10^{6}[/tex]  m/s

Explanation:

given data

wavelength [tex]\lambda 1[/tex] = 121.6 nm

wavelength [tex]\lambda 2[/tex] = 124.2 nm

solution

we get here first change in wavelength that is

[tex]\triangle \lambda[/tex] = 124.2 nm - 121.6 nm    ...............1

[tex]\triangle \lambda[/tex] = 2.6 nm  

we get here velocity of direction of the Earth that is express as

[tex]\frac{\triangle \lambda }{\lambda 1} =\frac{v}{c}[/tex]     ............2

put here value we get v

v = [tex]\frac{3\times 10^8\times 2.6}{121.6}[/tex]    

v = 6.4 × [tex]10^{6}[/tex]  m/s