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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 $13 imes 10^9$ years ago - Edexcel - A-Level Physics - Question 15 - 2023 - Paper 2

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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 $13 imes 10^9$ y... show full transcript

Worked Solution & Example Answer: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 $13 imes 10^9$ years ago - Edexcel - A-Level Physics - Question 15 - 2023 - Paper 2

Step 1

Derive the equation $T = \frac{1}{H_0}$, where $T$ is the age of the universe.

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Answer

To derive the equation, we start with Hubble's Law, which states that the velocity of a galaxy (vv) is proportional to its distance (dd) from us, represented as:

v=H0dv = H_0 d

Here, H0H_0 is the Hubble constant. We can express the distance in terms of time:

d=vtd = v t

Combining these equations gives:

v=H0vtv = H_0 vt

Rearranging gives:

t=1H0t = \frac{1}{H_0}

Thus, we derive the equation for the age of the universe.

Step 2

State one assumption made in your derivation.

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One assumption is that the expansion of the universe has been uniform, meaning the rate of expansion (Hubble's constant) has remained constant over time.

Step 3

Deduce whether the observation by the JWST leads to a value of $H_0$ within the accepted range.

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Answer

Using the relation between parsecs and kilometers:

1 Mpc = 3.1imes10163.1 imes 10^{16} m = 3.24imes1063.24 imes 10^6 light years.

To convert Hubble's constant from km/s/Mpc to SI units:

H0=(60 to 80) km/s3.1×1016 mH_0 = \frac{(60 \text{ to } 80) \text{ km/s}}{3.1 \times 10^{16} \text{ m}}

This calculation must be done to determine if the value obtained is within the accepted range of 608060-80 km/s/Mpc.

Step 4

Calculate the wavelength of light emitted by Maisie.

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Answer

The wavelength detected from Maisie is given by the formula:

λemitted=λobserved×(1+z)\lambda_{emitted} = \lambda_{observed} \times (1 + z)

Substituting the values:

λemitted=4.0×106 m×(1+14)=4.0×106 m×15=6.0×105 m\lambda_{emitted} = 4.0 \times 10^{-6} \text{ m} \times (1 + 14) = 4.0 \times 10^{-6} \text{ m} \times 15 = 6.0 \times 10^{-5} \text{ m}

Thus, the wavelength emitted is 6.0×1056.0 \times 10^{-5} m.

Step 5

Explain why the light emitted by Maisie arrives at the telescope as infrared.

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Answer

The light emitted by Maisie is redshifted due to the high value of zz. The expansion of the universe stretches the wavelength of the light, moving it from its original emitted frequency into the infrared region of the electromagnetic spectrum.

Step 6

Deduce whether this detector can detect the light from Maisie.

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Answer

The work function of the detector is 0.30 eV. To assess whether it can detect the light from Maisie, we must calculate the energy of the photons from Maisie's emitted light:

Using:

E=hcλE = \frac{hc}{\lambda}

where h=6.63×1034h = 6.63 \times 10^{-34} J·s and c=3.0×108c = 3.0 \times 10^8 m/s, we find the energy corresponding to the wavelength 6.0×1056.0 \times 10^{-5} m:

E=6.63×1034×3.0×1086.0×105=3.31×1019 J=3.31×10191.6×10192.07exteVE = \frac{6.63 \times 10^{-34} \times 3.0 \times 10^8}{6.0 \times 10^{-5}} = 3.31 \times 10^{-19} \text{ J} = \frac{3.31 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 2.07 ext{ eV}

Since 2.07 eV is greater than 0.30 eV, the detector can detect the light from Maisie.

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