A-level Physics (Advancing Physics)/Doppler Effect/Worked Solutions

1. M31 (the Andromeda galaxy) is approaching us at about 120kms−1. What is its red-shift?

${\displaystyle z={\frac {v_{s}}{c}}={\frac {-120000}{300000000}}=-0.4\times 10^{-3}}$

The minus sign is important! Andromeda is blue-shifted!

2. Some light from M31 reaches us with a wavelength of 590 nm. What is its wavelength, relative to M31?

${\displaystyle -0.0004={\frac {\Delta \lambda }{\lambda _{0}}}={\frac {\lambda -\lambda _{0}}{\lambda _{0}}}={\frac {\lambda }{\lambda _{0}}}-1={\frac {590\times 10^{-9}}{\lambda _{0}}}-1}$

${\displaystyle 0.9996={\frac {590\times 10^{-9}}{\lambda _{0}}}}$

${\displaystyle \lambda _{0}={\frac {590\times 10^{-9}}{0.9996}}=590.23{\mbox{ nm}}}$

3. Some light has a wavelength, relative to M31, of 480 nm. What is its wavelength, relative to us?

${\displaystyle 0.9996={\frac {\lambda }{\lambda _{0}}}={\frac {\lambda }{480\times 10^{-9}}}}$

${\displaystyle \lambda =0.9996\times 480\times 10^{-9}=479.808{\mbox{ nm}}}$

4. A quasar emits electromagnetic radiation at a wavelength of 121.6 nm. If, relative to us, this wavelength is red-shifted 0.2 nm, what is the velocity of recession of the quasar?

${\displaystyle {\frac {v_{s}}{c}}={\frac {\Delta \lambda }{\lambda _{0}}}}$

${\displaystyle {\frac {v_{s}}{3\times 10^{8}}}={\frac {0.2}{121.6}}=0.00164}$

${\displaystyle v_{s}=3\times 10^{8}\times 0.00164=493{\mbox{ kms}}^{-1}}$

However, this is about as high a velocity as we can use the classical Doppler effect for.