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Date May 2017 Marks available 2 Reference code 17M.2.HL.TZ2.5
Level Higher level Paper Paper 2 Time zone Time zone 2
Command term Write down 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]
a.

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

\(_2^4\alpha \)

OR

\({}_2^4{\text{He}}\)

\({}_{86}^{222}{\text{Rn}}\)

 

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

a.

ALTERNATIVE 1

6 days is 5.18 x 105 s

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

OR

A = 0.9999927 Aor 0.9999927 \(\lambda \)N0

OR

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

OR

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 = nNA = 3.7 x 1020

OR

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\({e^{ - \lambda t}}\)

N0n x NA = 3.7 x 1020

alpha particles emitted «= number of atoms disintegrated = N – N0N0\(\left( {1 - {e^{ - \lambda  \times 6 \times 24 \times 3600}}} \right)\) 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 «\(\frac{{291}}{{1.3 \times {{10}^{ - 5}}}}\)»

OR

p = 2.7 x 1015 x 1.3 x 10–23 x «\(\frac{{291}}{{1.3 \times {{10}^{ - 5}}}}\)»

0.83 or 0.84 «Pa»

 

Allow ECF for 2.7 x 1015 from (b)(ii).

c.ii.

Examiners report

[N/A]
a.
[N/A]
b.i.
[N/A]
b.ii.
[N/A]
c.i.
[N/A]
c.ii.

Syllabus sections

Core » Topic 7: Atomic, nuclear and particle physics » 7.1 – Discrete energy and radioactivity
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