Question

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 1

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

$\frac{V_r}{c} = \frac{\Delta \lambda}{\lambda_0}$

Here,

$V_r$ = Radial Velocity

c = Speed of light

$\Delta \lambda$ = Shift in wavelength

$\lambda_0$ = Laboratory wavelength of spectral line

Rearrange for $V_r$,

$V_r = \frac{\Delta \lambda}{\lambda_0} c$

Find shift in wavelength, $\Delta \lambda$

$\Delta \lambda = |485.5nm - 486.1nm|$

$\Delta \lambda = 0.6nm$

Replacing our values we have then,

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

$V_r = 370000m/s$

Therefore the radial velocity of star is $3.7*10^5$m/s

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