Two stable isotopes of helium are \( ^4_2He \) and \( ^3_2He \) - AQA - A-Level Physics - Question 1 - 2022 - Paper 1

The exchange particle responsible for this decay is the W- boson. This interaction involves a neutron changing into a proton, emitting a ( \beta^- ) particle (electron) and an antineutrino. This process is known as beta decay.

The evidence for the presence of helium is provided by the distinct emission lines labelled A to F in the emission spectrum of helium. These lines correspond to specific wavelengths emitted by helium atoms.

Contrarily, lines C and D can also correspond to those from hydrogen and sodium spectra. If sodium was present, we would expect additional lines that match sodium's emission spectrum, which seems to be absent.

Thus, while there may be overlapping features, the specific pattern and prominence of helium's lines A to F suggest that helium's absorption features can't solely be attributed to sodium or hydrogen.

To calculate the energy change related to spectral line E, we use the formula:

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

Where:

• ( h ) is Planck's constant (6.626 x 10^{-34} Js)
• ( c ) is the speed of light (3.0 x 10^8 m/s)
• ( \lambda ) is the wavelength of line E (assumed here to be approximately 586 nm as inferred from Figure 1).

Converting ( \lambda ) to meters: ( \lambda = 586 \times 10^{-9} m )

Now substituting in: $E = \frac{(6.626 \times 10^{-34} Js)(3.0 \times 10^8 m/s)}{586 \times 10^{-9} m}$

This will yield the energy in joules, which can then be converted to eV (1 eV = 1.602 x 10^{-19} J).