IB DP Physics Mock Exam HL Paper 2 Set 5 - 2025 Syllabus

(a)(i)
In alpha decay, proton number is conserved:
²²⁶₈₈Ra → ²²²_ZRn + ⁴₂He
\( 88 = Z + 2 \Rightarrow Z = 86 \).
Answer: \( \boxed{86} \)

(a)(ii)
Mass defect:
\( \Delta m = 226.02540 – 222.01757 – 4.00260 = 0.00523 \, \text{u} \).
Energy released:
\( E = \Delta m \times 931.5 = 0.00523 \times 931.5 \approx 4.9 \, \text{MeV} \).
Answer: \( \boxed{\approx 5 \, \text{MeV}} \)

(a)(iii)
Momentum is conserved, so the alpha particle and radon nucleus have equal momentum.
Since \( K \propto \frac{1}{m} \):
Fraction of energy carried by α:
\( \frac{222}{226} \approx 0.98 \).
Answer: \( \boxed{\text{About } 98\%} \)

(b)(i)
• Most alpha particles passed through the foil with little or no deflection.
• A small fraction were scattered through large angles, with some rebounding backwards.

(b)(ii)
These results show that the atom contains a very small, dense, positively charged nucleus that holds most of the mass, while the rest of the atom is largely empty space.

(b)(iii)
At the distance of closest approach, kinetic energy equals electrostatic potential energy:
\( K = \frac{1}{4\pi\varepsilon_0}\frac{(2e)(79e)}{r} \).
Substitution gives:
\( r \approx 4.7 \times 10^{-14} \, \text{m} \).
Answer: \( \boxed{r_{\text{min}} \approx 4.7 \times 10^{-14} \, \text{m}} \)

(b)(iv)
Nuclear radius is given by \( R = R_0 A^{1/3} \).
\( \frac{R_{\text{Au}}}{R_{\text{Al}}} = \left(\frac{197}{27}\right)^{1/3} \approx 1.9 \).
Answer: \( \boxed{\approx 1.9} \)