| Date | May 2017 | Marks available | 1 | Reference code | 17M.2.HL.TZ2.5 |
| Level | Higher level | Paper | Paper 2 | Time zone | Time zone 2 |
| Command term | Outline | 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.
Deduce that the activity of the radium-226 is almost constant during the experiment.
Show that about 3 x 1015 alpha particles are emitted by the radium-226 in 6 days.
The wall of cylinder A is made from glass. Outline why this glass wall had to be very thin.
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.
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.
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 A0 or 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.
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\({e^{ - \lambda t}}\)
N0 = n x NA = 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
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.
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).
