DP Physics (first assessment 2025)

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Question 23M.2.HL.TZ1.i

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Date May 2023 Marks available [Maximum mark: 2] Reference code 23M.2.HL.TZ1.i
Level HL Paper 2 Time zone TZ1
Command term Show that Question number i Adapted from N/A
i.
[Maximum mark: 2]
23M.2.HL.TZ1.i

In an experiment, alpha particles of initial kinetic energy 5.9 MeV are directed at stationary nuclei of lead (Pb82207). Show that the distance of closest approach is about 4 × 10−14 m.

[2]

Markscheme

2e AND 82e seen

OR

3.2 × 10−19 «C» AND 1.312 × 10−17 «C» seen ✓

d = 8.99×109×(2e)(82e)5.9×106×e = 3.998 × 10−14 ≈ 4 × 10−14 «m» ✓

Must see either clear substitutions or answer to at least 4 s.f. for MP2.

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