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Date	May 2017	Marks available	3	Reference code	17M.2.HL.TZ2.5
Level	Higher level	Paper	Paper 2	Time zone	2
Command term	Show that	Question number	5	Adapted from	N/A

Question

The first scientists to identify alpha particles by a direct method were Rutherford and Royds. They knew that radium-226 ( ${}_{86}^{226}{\text{Ra}}$ ) decays by alpha emission to form a nuclide known as radon (Rn).

At the start of the experiment, Rutherford and Royds put 6.2 x 10^–4 mol of pure radium-226 in a small closed cylinder A. Cylinder A is fixed in the centre of a larger closed cylinder B.

The experiment lasted for 6 days. The decay constant of radium-226 is 1.4 x 10^–11 s^–1.

At the start of the experiment, all the air was removed from cylinder B. The alpha particles combined with electrons as they moved through the wall of cylinder A to form helium gas in cylinder B.

Write down the nuclear equation for this decay.

[2]

Deduce that the activity of the radium-226 is almost constant during the experiment.

[2]

b.i.

Show that about 3 x 10¹⁵ alpha particles are emitted by the radium-226 in 6 days.

[3]

b.ii.

The wall of cylinder A is made from glass. Outline why this glass wall had to be very thin.

[1]

c.i.

The experiment was carried out at a temperature of 18 °C. The volume of cylinder B was 1.3 x 10^–5 m³ and the volume of cylinder A was negligible. Calculate the pressure of the helium gas that was collected in cylinder B over the 6 day period. Helium is a monatomic gas.

[3]

c.ii.

Markscheme

$_2^4\alpha$

${}_2^4{\text{He}}$

${}_{86}^{222}{\text{Rn}}$

These must be seen on the right-hand side of the equation.

ALTERNATIVE 1

6 days is 5.18 x 10⁵ s

activity after 6 days is ${A_0}{e^{ - 1.4 \times {{10}^{ - 11}} \times 5.8 \times {{10}^5}}} \approx {A_0}$

A = 0.9999927 A₀or 0.9999927 $\lambda$ N₀

states that index of e is so small that $\frac{A}{{{A_0}}}$ is ≈ 1

A – A₀ ≈ 10^–15 «s^–1»

ALTERNATIVE 2
shows half-life of the order of 10¹¹ s or 5.0 x 10¹⁰ s

converts this to year «1600 y» or days and states half-life much longer than experiment compared to experiment

Award [1 max] if calculations/substitutions have numerical slips but would lead to correct deduction.

eg: failure to convert 6 days to seconds but correct substitution into equation will give MP2.

Allow working in days, but for MP1 must see conversion of $\lambda$ or half-life to day^–1.

b.i.

ALTERNATIVE 1

use of A = $\lambda$ N₀

conversion to number of molecules = nN_A = 3.7 x 10²⁰

initial activity = 5.2 x 10⁹ «s^–1»

number emitted = (6 x 24 x 3600) x 1.4 x 10^–11 x 3.7 x 10²⁰ or 2.7 x 10¹⁵ alpha particles

ALTERNATIVE 2
use of N = N₀ ${e^{ - \lambda t}}$

N₀ = n x N_A = 3.7 x 10²⁰

alpha particles emitted «= number of atoms disintegrated = N – N₀ =» N₀ $\left( {1 - {e^{ - \lambda \times 6 \times 24 \times 3600}}} \right)$ or 2.7 x 10¹⁵ alpha particles

Must see correct substitution or answer to 2+ sf for MP3

b.ii.

alpha particles highly ionizing
OR
alpha particles have a low penetration power
OR
thin glass increases probability of alpha crossing glass
OR
decreases probability of alpha striking atom/nucleus/molecule

Do not allow reference to tunnelling.

c.i.

conversion of temperature to 291 K

p = 4.5 x 10^–9 x 8.31 x « $\frac{{291}}{{1.3 \times {{10}^{ - 5}}}}$ »