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 590 nm. 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 480 nm. 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 121.6 nm. If, relative to us, this wavelength is red-shifted 0.2 nm, 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.