A-level
PHYSICS
Paper 1
2
*02*
IB/M/Jun22/7408/1
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outside the Section A box
Answer all questions in this section.
0 1 Two stable isotopes of helium are 4
2
He and 3
2
He.
0 1 . 1 An atom of 4
...
A-level
PHYSICS
Paper 1
2
*02*
IB/M/Jun22/7408/1
Do not write
outside the Section A box
Answer all questions in this section.
0 1 Two stable isotopes of helium are 4
2
He and 3
2
He.
0 1 . 1 An atom of 4
2
He is produced in a rock that contains uranium. It is produced following
the radioactive decay of a 238
92U atom. The decay also creates an atom of
thorium (Th).
Write an equation for the decay of 238
92U.
[2 marks]
238
92U →
0 1 . 2 A 3
2
He nucleus can be produced by the decay of a tritium nucleus 3
1
H.
State and explain which exchange particle is responsible for this decay.
[2 marks]
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outside the
box
3
*03*
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Helium was discovered by analysing the light in the box absorption spectrum of the Sun.
Figure 1 shows the positions of the brightest lines, labelled A to F, in the emission
spectrum of helium. The brightest lines in the emission spectra of sodium and
hydrogen are also shown.
Figure 1
0 1 . 3 Before helium was identified, some scientists suggested that the lines of the helium
spectrum seen in the absorption spectrum of the Sun were due to the presence of
sodium and hydrogen.
Discuss, with reference to the lines A to F in Figure 1, the evidence for and against
this suggestion.
[2 marks]
Question 1 continues on the next page
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0 1 box . 4 Calculate, in eV, the change in energy level responsible for the spectral line labelled E
in Figure 1.
[3 marks]
change in energy level = eV
0 1 . 5 Explain, with reference to the processes within an atom, the difference between an
emission spectrum and an absorption spectrum.
[3 marks]
12
5
*05*
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0 2 box Carbon-14 decays into nitrogen-14 with the release of a beta (β−) particle and an
antineutrino ( e v ).
0 2 . 1 State the change of quark character in β− decay.
[1 mark]
0 2 . 2 Figure 2 shows the distribution of kinetic energies of β− particles from the decay of
carbon-14.
Figure 2
Explain how Figure 2 supports the existence of the antineutrino.
[2 marks]
Question 2 continues on the next page
6
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box The existence of the antineutrino was confirmed by experiments in which
antineutrinos interact with protons. The equation for this interaction is:
e v + p → e+
+ X
0 2 . 3 Identify particle X.
[1 mark]
0 2 . 4 The positron released in this interaction is annihilated when it encounters an electron.
A pair of gamma photons is then produced.
Particle X can be absorbed by a nucleus. This produces another gamma ray.
Table 1 contains data for three gamma photons detected during an
antineutrino–proton interaction experiment.
Table 1
Gamma photon Photon energy / J
G1 5.0 × 10−14
G2 6.6 × 10−14
G3 1.0 × 10−13
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