The wavelength of a particular spectrum line is 6000 A in the laboratory. The wavelength of the same line is observed to be 6020 A in the spectrum of a star. Then which of the following is correct?

CONCEPT:

• Doppler shift is also known as the Doppler effect is defined as the change in the wavelength or frequency of the waves with respect to the observer who is in motion relative to the wave source.
• The phenomenon was described in the year 1842 by the Austrian physicist Christian Doppler.

CALCULATION:

Given λ_A = 6020 Å and λ_s = 6000 Å

• Doppler's shift is given by,

\(⇒ Δλ=\frac{vλ_{S}}{c}\)     -----(1)

Where, Δλ = doppler shift(Å), v = velocity(m/s), λ_S = wavelength of source(Å), and c = speed of light = 3×108 m/s

\(⇒ Δλ=\lambda_{S}-\lambda_{A}\)

\(⇒ Δλ=6000-6020\)

\(⇒ Δλ=-20​​Å\)     -----(2)

By equation 1 and equation 2,

\(\Rightarrow -20​​Å=\frac{v\times6000Å}{3\times10^{8}}\)

\(\Rightarrow -20​​Å=\frac{v\times6000Å}{3\times10^{8}}\)

\(\Rightarrow v=-10^6m/s\)

• The negative sign shows that the star is moving towards the earth. Hence, option 3 is correct.