Paper 4 Atomic Structure Answers

Q1 Measuring an Emission-Line Wavelength

1. $600\mathrm{m}{\mathrm{m}}^{-1}=6.00\times 10^5{\mathrm{m}}^{-1}$, hence $d=1/(6.00\times 10^5)=1.67\times 10^{-6}\mathrm{m}$. [2]
2. $\bar{\theta }=19.5^{\circ }$ and $\Delta \theta =(19.6-19.4)/2=0.1^{\circ }$. [2]
3. $\lambda =d\sin 19.5^{\circ }=5.56\times 10^{-7}\mathrm{m}$ ($556\mathrm{nm}$). [2]
4. ${\lambda }_{\max }=d\sin 19.6^{\circ }\approx 5.59\times 10^{-7}\mathrm{m}$ and ${\lambda }_{\min }=d\sin 19.4^{\circ }\approx 5.54\times 10^{-7}\mathrm{m}$. Half their range is approximately $2.7\times 10^{-9}\mathrm{m}$, so quote $\lambda \approx (5.56\pm 0.03)\times 10^{-7}\mathrm{m}$. Accept consistent rounding. [3]
5. $E=hc/\lambda \approx 3.58\times 10^{-19}\mathrm{J}$. [2]
6. Measure symmetrically on both sides of the central maximum and average, use a larger-order maximum where resolved, or repeat more readings [1]. Avoid contact with the high-voltage discharge-tube supply or switch off before adjusting it [1].

Q2 Rutherford Experiment Evaluation

1. To reduce multiple scattering and allow alpha particles to pass through. [1]
2. Alpha particles are massive, positively charged and energetic, so they travel mostly straight but can be strongly repelled near nuclei. [2]
3. Scattering gives nuclear size/charge concentration evidence but does not itself explain stable electron arrangements or quantised spectra. [2]