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Date May 2017 Marks available 1 Reference code 17M.2.SL.TZ2.4
Level Standard level Paper Paper 2 Time zone Time zone 2
Command term Outline Question number 4 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).

Write down the missing values in the nuclear equation for this decay.

[1]
a.

Rutherford and Royds put some pure radium-226 in a small closed cylinder A. Cylinder A is fixed in the centre of a larger closed cylinder B.

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.

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

[1]
b.

Rutherford and Royds expected 2.7 x 1015 alpha particles to be emitted during the experiment. 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.

[3]
c.

Rutherford and Royds identified the helium gas in cylinder B by observing its emission spectrum. Outline, with reference to atomic energy levels, how an emission spectrum is formed.

[3]
d.

The work was first reported in a peer-reviewed scientific journal. Outline why Rutherford and Royds chose to publish their work in this way.

[1]
e.

Markscheme

222 AND 4

 

Both needed.

a.

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

b.

conversion of temperature to 291 K

p = 4.5 x 10–9 x 8.31 x «\(\frac{{2.91}}{{1.3 \times {{10}^{ - 5}}}}\)»

OR

p = 2.7 x 1015 x 1.38 x 10–23 x «\(\frac{{2.91}}{{1.3 \times {{10}^{ - 5}}}}\)»

0.83 or 0.84 «Pa»

 

c.

electron/atom drops from high energy state/level to low state

energy levels are discrete

wavelength/frequency of photon is related to energy change or quotes Ehf or E = \(\frac{{hc}}{\lambda }\)

and is therefore also discrete

d.

peer review guarantees the validity of the work
OR
means that readers have confidence in the validity of work

 

OWTTE

e.

Examiners report

[N/A]
a.
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b.
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c.
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d.
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e.

Syllabus sections

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