NESA Physics Properties of the Nucleus
7 sample questions with marking guides and sample answers · Avg. score: 71.9%
A scientist has two unlabelled sources of radiation. One source emits alpha particles and the other emits beta particles.
Outline TWO methods that could be used to determine which source is the alpha emitter, and which source is the beta emitter.
Reveal Answer
Pass the radiation through a thin material to a detector. Observe any changes in counts. The beta source count will be reduced by an amount less than the alpha source count.
Pass each type of radiation through a magnetic field and observe any differences in deflection and observe deflection in opposite directions.
| Descriptor | Marks |
|---|---|
Outlines TWO methods and subsequent observations that could be used to identify each source as either alpha or beta emitters | 3 |
| 2 |
Provides some relevant information | 1 |
None of the above | 0 |
Caesium-137 has a half-life of 30 years.
What mass of caesium-137 will remain after 90 years, if the initial mass was 120 g?
4 g
15 g
40 g
60 g
Reveal Answer
4 g
This incorrect answer comes from dividing the initial mass by the half-life (), rather than calculating the exponential decay over 3 half-lives.
15 g
Since 90 years is exactly 3 half-lives (), the initial mass is halved 3 times. The remaining mass is .
40 g
This incorrect answer comes from dividing the initial mass by the number of half-lives (), instead of dividing by .
60 g
This is the mass that would remain after only 1 half-life (30 years), not the 3 half-lives (90 years) specified in the question.
A neutron is absorbed by a nucleus, .
The resulting nucleus undergoes alpha decay, producing lithium-7.
What is nucleus ?
Boron-10
Boron-11
Lithium-6
Lithium-10
Reveal Answer
Boron-10
Working backwards, Lithium-7 () plus an alpha particle () gives an intermediate nucleus of Boron-11 (). Since this was formed by absorbing a neutron (), the original nucleus must be Boron-10 ().
Boron-11
If Boron-11 absorbed a neutron, it would form Boron-12. An alpha decay from Boron-12 would produce Lithium-8, not Lithium-7.
Lithium-6
If Lithium-6 absorbed a neutron, it would form Lithium-7 directly, but the question states that Lithium-7 is produced via alpha decay after the neutron absorption.
Lithium-10
If Lithium-10 absorbed a neutron, it would form Lithium-11. An alpha decay from Lithium-11 would produce Hydrogen-7, not Lithium-7.
Nucleus X has a greater binding energy than nucleus Y.
What can be deduced about X and Y?
X is more stable than Y.
Y is more stable than X.
X has a greater mass defect than Y.
Y has a greater mass defect than X.
Reveal Answer
X is more stable than Y.
Stability is determined by binding energy per nucleon, not total binding energy, so we cannot determine which nucleus is more stable without knowing their nucleon numbers.
Y is more stable than X.
Stability depends on binding energy per nucleon, which cannot be determined from total binding energy alone.
X has a greater mass defect than Y.
Binding energy is directly proportional to mass defect according to the mass-energy equivalence equation . Therefore, a greater binding energy means a greater mass defect.
Y has a greater mass defect than X.
Since nucleus X has a greater binding energy, it must have a greater mass defect than Y, not the other way around.
Carbon-14 undergoes nuclear decay to nitrogen-14.
List the two types of particles whose total number must be conserved in this reaction.
Reveal Answer
Baryons
Leptons
| Descriptor | Marks |
|---|---|
Identifies baryons and leptons | 1 |
A pure sample of polonium-210 undergoes alpha emission to produce the stable isotope lead-206.
The half-life of polonium-210 is 138 days.
At the end of 276 days, what is the ratio of polonium-210 atoms to lead-206 atoms in the sample?
1:4
1:3
1:2
1:1
Reveal Answer
1:4
This is incorrect. While of the original polonium-210 remains, the remaining has decayed into lead-206, making the ratio , not .
1:3
This is correct. 276 days is exactly two half-lives (). After two half-lives, of the original polonium-210 remains, and has decayed into lead-206, resulting in a ratio.
1:2
This is incorrect. A ratio would mean of the sample is polonium-210 and is lead-206, which does not correspond to an integer number of half-lives.
1:1
This is incorrect. A ratio occurs after exactly one half-life (138 days), when half of the polonium-210 has decayed into lead-206.
Gamma radiation is often used to treat cancerous tumours. The energy of a gamma photon emitted by radioactive cobalt-60 is 1.33 MeV.
Which one of the following is closest to the frequency of the gamma radiation?
Reveal Answer
This is incorrect because it simply takes the numerical value of the energy in eV () and assigns it the unit of frequency (Hz) without applying Planck's equation.
This is correct. Using Planck's equation , the frequency is . Using the energy in eV () and Planck's constant in eV s (), we get .
This is incorrect. The value is off by a factor of 10, which likely results from a simple arithmetic or unit conversion error when calculating .
This is incorrect because it results from dividing the energy in eV () by Planck's constant in Joules seconds (). Units must be consistent (either both in eV or both in Joules) before dividing.