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 the decay of a tritium nucleus ($^3_1H$) is the W boson. This is a weak interaction where a neutron converts into a proton, emitting a beta particle (electron) and an antineutrino. The W boson mediates this change, facilitating the transformation while conserving charge.

The lines labeled A to F in Figure 1 represent the wavelengths at which helium absorbed light in the spectrum. The presence of these lines indicates the specific energy levels of electrons in helium that can absorb photons of exact energies corresponding to their transitions. The absence of such lines in the spectra of sodium and hydrogen further supports the uniqueness of the helium spectrum. While sodium and hydrogen have their own absorption lines due to their respective electron transitions, the spectral emissions specifically related to helium demonstrate distinct energy level configurations.

The wavelength corresponding to line E in Figure 1 is approximately 580 nm. The energy of a photon can be calculated using the formula:

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

Where:

• $h$ is Planck’s constant ($6.626 \times 10^{-34} J s$)
• $c$ is the speed of light ($3.0 \times 10^8 m/s$)
• $\lambda$ is the wavelength in meters (580 nm = $580 \times 10^{-9} m$)

Substituting the values:

$E = \frac{(6.626 \times 10^{-34})(3.0 \times 10^8)}{580 \times 10^{-9}}$

Calculating this gives an energy of approximately 3.41 x 10^-19 J. To convert this result into eV, divide by the charge of an electron (1.602 x 10^-19 C):

$E (eV) \approx \frac{3.41 \times 10^{-19}}{1.602 \times 10^{-19}} \approx 2.13 eV$