4. Light from the Sun is used to produce a visible spectrum - Scottish Highers Physics - Question 4 - 2015

The position of the hydrogen line differs in each spectrum due to the Doppler effect. As the galaxies move away from the Sun, the wavelengths of the spectral lines are redshifted, meaning the observed wavelengths are longer than they are in the stationary frame of reference. Consequently, the distant galaxy shows a greater redshift compared to the nearby galaxy.

To show that the redshift (z) is 0.098, we can use the formula:

$z = \frac{\lambda_{observed} - \lambda_{emitted}}{\lambda_{emitted}}$

Using the provided wavelengths, $\lambda_{observed} = 450 \text{ nm}$ and $\lambda_{emitted} = 410 \text{ nm}$:

$z = \frac{450 - 410}{410} = \frac{40}{410} = 0.098$

Using the redshift calculated earlier (z = 0.098), we can estimate the distance (d) to the galaxy using Hubble's law:

$v = H_0 d$

Where:

• v is the velocity of the galaxy, given by: $v = cz$, with c being the speed of light ($c = 3.00 \times 10^8 m/s$).

Therefore: $v = (0.098)(3.00 \times 10^8) \approx 2.94 \times 10^7 m/s$

Hubble's constant (H₀) is typically taken as approximately 70 km/s/Mpc. Converting this to m/s/Mpc: $H_0 \approx 2.27 \times 10^{-18} s^{-1}$

Substituting back into Hubble's law gives us:

$d = \frac{v}{H_0} \approx \frac{2.94 \times 10^7}{2.27 \times 10^{-18}} \approx 1.3 \times 10^{25} m \approx 4.2 \times 10^{21} light-years$