# 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?

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

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

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

$-0.0004 = \frac{\Delta\lambda}{\lambda_0} = \frac{\lambda - \lambda_0}{\lambda_0} = \frac{\lambda}{\lambda_0} - 1 = \frac{590 \times 10^{-9}}{\lambda_0} - 1$

$0.9996 = \frac{590 \times 10^{-9}}{\lambda_0}$

$\lambda_0 = \frac{590 \times 10^{-9}}{0.9996} = 590.23\mbox{ nm}$

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

$0.9996 = \frac{\lambda}{\lambda_0} = \frac{\lambda}{480 \times 10^{-9}}$

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

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

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

$\frac{v_s}{3 \times 10^8} = \frac{0.2}{121.6} = 0.00164$

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

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