1. An atom of $^{238}_{92}U$ undergoes decay as shown below: $$^{238}_{92}U \rightarrow ^{234}_{90}Th + ^4_2He$$ 2 - AQA - A-Level Physics - Question 1 - 2022 - Paper 1

The exchange particle responsible for the decay of a $^{3}_{1}H$ nucleus to produce a $^{3}_{2}He$ nucleus is the W boson. This is a weak interaction, involving the weak force, which allows for the transformation of a neutron into a proton.

Regarding the evidence for and against the suggestion that the lines in the absorption spectrum form due to helium:

• Lines C in Figure 1 correspond to both hydrogen and helium spectra, supporting the presence of helium in the Sun.
• Conversely, line E shows consistency with sodium and helium spectra, indicating possible overlap.
• Notably, line D is absent in the spectrum of sodium, confirming that some lines can be derived from helium exclusively. Thus, there are arguments both for and against the presence of helium, but the patterns in the absorption spectrum suggest its involvement.

To calculate the energy for the line labelled E, we first identify its wavelength. It falls within the range of 580 nm to 590 nm: Using the formula: $E = \frac{hc}{\lambda}$ Where:

• $h = 6.626 \times 10^{-34} J·s$
• $c = 3 \times 10^8 m/s$. Thus, converting to eV: The wavelength of line E is approximately 590 nm, then: $E = \frac{(6.626 \times 10^{-34} J·s)(3 \times 10^8 m/s)}{590 \times 10^{-9} m}$ This yields an energy approximately in the range of 2.1 eV.