A particle P moves with simple harmonic motion.

(i) State, with reference to the motion of P, what is meant by simple harmonic motion.

(ii) State the phase difference between the displacement and the velocity of P.

The diagram shows four spectral lines in the visible line emission spectrum of atomic hydrogen.

(i) Outline how such a spectrum may be obtained in the laboratory.

(ii) Explain how such spectra give evidence for the existence of discrete atomic energy levels.

The energies of the principal energy levels in atomic hydrogen measured in eV are given by the expression

\({E_n} =  - \frac{{13.6}}{{{n^2}}}\) where n=1, 2, 3 ..........

The visible lines in the spectrum correspond to electron transitions that end at n=2.

(i) Calculate the energy of the level corresponding to n=2.

(ii) Show that the spectral line of wavelength λ=485nm is the result of an electron transition from n=4.

The alpha particles and gamma rays produced in radioactive decay have discrete energy spectra. This suggests that nuclei also possess discrete energy levels. However, beta particles produced in radioactive decay have continuous energy spectra. Describe how the existence of the antineutrino accounts for the continuous nature of beta spectra.

(i) the acceleration (of a particle/P) is (directly) proportional to displacement;
and is directed towards equilibrium/in the opposite direction to displacement;
Do not accept “directed towards the centre”.

(ii) \(\frac{\pi }{2}\)/90°/quarter of a period;

(i) light from a hydrogen discharge tube/hot hydrogen gas/ hydrogen tube with potential difference across it;
is passed onto a prism/diffraction grating;
and then is observed on a screen/through a telescope;
Accept good labelled diagram for explanation of any marking point.

(ii) each wavelength corresponds to the energy of the photon emitted;
when an electron makes a transition from a higher to lower energy level;
since only discrete wavelengths/finite number of wavelengths are present, then only discrete energy levels are present / OWTTE;

(i) −3.40 eV;
Award [0] for omitted negative sign.

(ii) energy difference between levels\( = \frac{{hc}}{{\lambda e}} = \frac{{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{4.85 \times {{10}^{ - 7}} \times 1.6 \times {{10}^{ - 19}}}}\);
=2.55eV;
\(\left[ {3.40 - 2.55} \right] = 0.85 = \frac{{13.6}}{{{n^2}}}\) to give n²=16;
n=4;
Award [3] for reversed argument.

the total emitted energy is shared between the electron and the antineutrino;
the energy/velocity can be shared/distributed in an infinite number of ways / OWTTE;