The photograph below was taken by the James Webb Space Telescope (JWST) and shows a group of galaxies that formed shortly after the big bang, about 13 × 10^9 years ago - Edexcel - A-Level Physics - Question 15 - 2023 - Paper 2

One key assumption made is that the expansion of the universe has occurred uniformly over time, implying that the Hubble constant has remained constant throughout the age of the universe.

To deduce the value of the Hubble constant H₀, we first convert the parsec to kilometers:

1 parsec (pc) = 3.1 × 10^16 m = 3.1 × 10^13 km

The acceptable range for H₀ is (60–80) km/s/Mpc. Substituting the conversion factors:

• For 1 year = 3.16 × 10^7 s, we find the velocities:
• If v = 60 km/s/Mpc, then

$H₀ = \frac{60 km/s}{1.0 Mpc} \approx 1.89 × 10^{-18} s^{-1}$

• If v = 80 km/s/Mpc, then

$H₀ = \frac{80 km/s}{1.0 Mpc} \approx 2.66 × 10^{-18} s^{-1}$

Given these calculations, the Hubble constant from the JWST observations falls within this range.

The wavelength emitted by Maisie can be calculated using the redshift formula:

$\lambda_{observed} = \lambda_{emitted} (1 + z)$

Here, ( \lambda_{observed} = 4.0 × 10^{-10} m ) and ( z = 14 ):

This results in:

$4.0 × 10^{-10} = \lambda_{emitted} (1 + 14)$

Solving for ( \lambda_{emitted} ):

$\lambda_{emitted} = \frac{4.0 × 10^{-10}}{15} = 2.67 × 10^{-11} m$

The light from Maisie arrives at the telescope as infrared due to the phenomenon of redshift. As the universe expands, the wavelengths of light from distant galaxies stretch, moving them into the infrared spectrum. Given Maisie's high redshift of 14, this significant stretching results in the originally emitted light shifting from visible or ultraviolet to infrared frequencies.

To deduce whether the infrared detector can detect the light from Maisie, we must compare the energy of the photons emitted by Maisie to the work function of the detector. The energy of the emitted photons can be calculated using:

$E = \frac{hc}{\lambda}$

Where:

• h is Planck's constant (6.626 × 10^{-34} J∙s)
• c is the speed of light (3.00 × 10^8 m/s)
• ( \lambda = 2.67 × 10^{-11} m )

Calculating:

$E = \frac{(6.626 × 10^{-34})(3 × 10^8)}{2.67 × 10^{-11}} \approx 7.43 × 10^{-14} J \approx 0.464 eV$

Since 0.464 eV is greater than the work function of 0.30 eV, the detector can indeed detect the light from Maisie.