In a laboratory, the Balmer-beta spectral line of hydrogen has a wavelength of 486.1 nm. If the line appears in a star's spectrum at 485.5 nm, what is the star's radial velocity (in km/s)

Answer :

cjmejiab

Use Doppler's formula to find the radial velocity of star.

[tex]\frac{V_r}{c} = \frac{\Delta \lambda}{\lambda_0}[/tex]

Here,

[tex]V_r[/tex] = Radial Velocity

c = Speed of light

[tex]\Delta \lambda[/tex] = Shift in wavelength

[tex]\lambda_0[/tex] = Laboratory wavelength of spectral line

Rearrange for [tex]V_r[/tex],

[tex]V_r = \frac{\Delta \lambda}{\lambda_0} c[/tex]

Find shift in wavelength, [tex]\Delta \lambda[/tex]

[tex]\Delta \lambda = |485.5nm - 486.1nm|[/tex]

[tex]\Delta \lambda = 0.6nm[/tex]

Replacing our values we have then,

[tex]V_r = \frac{0.6nm}{486.1nm}(3*10^8m/s)[/tex]

[tex]V_r = 370000m/s[/tex]

Therefore the radial velocity of star is [tex]3.7*10^5[/tex]m/s

In this case the symbol of [tex]\Delta \lambda[/tex] implies that the star is receding the observer and the wavelength turns to red, then is red shifted.

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