DP Physics (last assessment 2024)

Question 22N.2.HL.TZ0.6b.ii

Date	November 2022	Marks available	[Maximum mark: 3]	Reference code	22N.2.HL.TZ0.6b.ii
Level	HL	Paper	2	Time zone	TZ0
Command term	Show	Question number	b.ii	Adapted from	N/A

b.ii.

[Maximum mark: 3]

22N.2.HL.TZ0.6b.ii

Polonium-210 (Po-210) decays by alpha emission into lead-206 (Pb-206).

The following data are available.

Nuclear mass of Po-210 = 209.93676 u

Nuclear mass of Pb-206 = 205.92945 u

Mass of the alpha particle = 4.00151 u

(b.ii)

The polonium nucleus was stationary before the decay.

Show, by reference to the momentum of the particles, that the kinetic energy of the alpha particle is much greater than the kinetic energy of the lead nucleus.

[3]

Markscheme

ALTERNATIVE 1

energy ratio expressed in terms of momentum, e.g. $\frac{E_{α}}{E_{lead}} = \frac{p_{α}^{2} / 2 m_{α}}{p_{lead}^{2} / 2 m_{lead}}$ ✓

$p_{α} = p_{lead}$ hence $\frac{E_{α}}{E_{lead}} = \frac{m_{lead}}{m_{α}}$ ✓

$\frac{m_{lead}}{m_{α}} ≃ \frac{206}{4} = 51.5 \Rightarrow E_{α} = 51.5 E_{lead}$ «so $α$ has a much greater KE»

$m_{lead}$ «much» greater than $m_{alpha}$ «so $α$ has a much greater KE» ✓

ALTERNATIVE 2

alpha particle and lead particle have equal and opposite momenta ✓

so their velocities are inversely proportional to mass ✓

but KE ∝ v² «so $α$ has a much greater KE» ✓

Examiners report

Those who answered using the mass often did not get MP3 whereas those who converted to the number of particles or moles before the first calculation did, although that could be considered an unnecessary complication.

Syllabus sections

Core

Core » Topic 2: Mechanics » 2.3 – Work, energy, and power

Core » Topic 2: Mechanics » 2.4 – Momentum and impulse

Core » Topic 2: Mechanics

Question 22N.2.HL.TZ0.6b.ii

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