IBDP Physics- E.1 Structure of the atom - IB Style Questions For SL Paper 2 -FA 2025
Question
Atomic mass of Rn \(= 222.01757 \,\text{u}\)
Mass of \( \alpha \) particle \(= 4.00260 \,\text{u}\)
Most-appropriate topic codes (IB Physics 2025):
• E.1.3: Geiger–Marsden–Rutherford experiment — parts (b)(i), (b)(ii)
▶️ Answer/Explanation
(a)(i)
Alpha decay reduces the proton number by \( 2 \).
\( Z_{\text{Rn}} = 88 – 2 = \boxed{86} \)
(a)(ii)
Mass defect:
\( \Delta m = 226.02540 – 222.01757 – 4.00260 = 0.00523 \,\text{u} \)
Energy released:
\( E = 0.00523 \times 931.5 = 4.87 \,\text{MeV} \approx 5 \,\text{MeV} \)
(a)(iii)
Conservation of momentum gives equal momenta for the alpha particle and radon nucleus.
Using \( \text{KE} = \dfrac{p^{2}}{2m} \):
\( \dfrac{\text{KE}_{\alpha}}{\text{KE}_{\text{Rn}}} = \dfrac{m_{\text{Rn}}}{m_{\alpha}} \approx \dfrac{222}{4} \)
Fraction of energy carried by the alpha particle:
\( \dfrac{222}{226} \approx 0.98 \)
(b)(i)
Most alpha particles passed straight through the foil without deflection.
A small fraction were scattered through large angles or reflected backwards.
(b)(ii)
These results show that the atom is mostly empty space.
The large deflections imply a very small, dense, positively charged nucleus at the centre of the atom.