ANWSER
Question 1:
(a) (i) Calculate the energy released in the α decay of ²³⁸U → ²³⁴Th + ⁴He
The energy released (Q-value) is calculated using the mass defect and Einstein’s equation:
Q=(mparent−mdaughter−mα)c2Q = (m_{\text{parent}} – m_{\text{daughter}} – m_{\alpha})c^2
Given:
- Mass of ²³⁸U = 238.050788 u
- Mass of ²³⁴Th = 234.043593 u
- Mass of ⁴He (α-particle) = 4.002603 u
Mass defect:
Δm=(238.050788−234.043593−4.002603)\Delta m = (238.050788 – 234.043593 – 4.002603) Δm=0.004592u\Delta m = 0.004592 u
Since 1 u = 931.5 MeV,
Q=0.004592×931.5=4.28 MeVQ = 0.004592 \times 931.5 = 4.28 \text{ MeV}
(ii) What fraction of the mass of a single ²³⁸U is destroyed in the decay?
Fraction=Δmmparent=0.004592238.050788≈1.93×10−5\text{Fraction} = \frac{\Delta m}{m_{\text{parent}}} = \frac{0.004592}{238.050788} \approx 1.93 \times 10^{-5}
(iii) Although the fractional mass loss is large for a single nucleus, it is difficult to observe for an entire macroscopic sample of uranium. Why is this?
The total mass of macroscopic uranium samples is very large compared to the mass defect per decay. Since the decay process occurs at an atomic scale, the mass lost is negligible on a macroscopic scale and is only detectable using precise instruments.
(b) (i) Define the band of stability and relate it to the value of the n/z ratio.
The band of stability refers to the region on a neutron-to-proton (n/z) ratio graph where stable nuclides are found. Lighter elements have an n/z ratio of 1, while heavier elements require a higher neutron-to-proton ratio (around 1.5) to remain stable.
(ii) What are transuranium elements?
Transuranium elements are elements with atomic numbers greater than 92 (Uranium). They are synthetic and undergo radioactive decay. Examples include Neptunium (Np), Plutonium (Pu), and Curium (Cm).
(iii) Describe briefly the work of G.T. Seaborg in this area of nuclear chemistry research.
G.T. Seaborg contributed significantly to the discovery of transuranium elements and helped develop the actinide concept, which correctly positioned the actinides in the periodic table. He co-discovered elements like Plutonium (Pu), Americium (Am), and Curium (Cm).
(c) (i) Define the term Nuclear Coulombic Energy Barrier.
The Nuclear Coulombic Energy Barrier is the electrostatic repulsion energy between two positively charged nuclei that must be overcome for nuclear reactions like fusion or α-decay to occur.
(ii) Consider an alpha particle just in contact with a ²³⁹Pu nucleus. Calculate the Coulombic repulsion energy assuming that the distance between them is equal to the sum of their radii.
Given:
- 4πε₀ = 1.11265 × 10⁻¹⁰ J·m
- Charge of ²³⁹Pu = 94e
- Charge of α-particle = 2e
- Distance (sum of radii) = r = 1.2 × (239¹/³ + 4¹/³) fm
- Convert distance to meters and calculate Coulombic energy using:
E=(Z1e)(Z2e)4πε0rE = \frac{(Z_1 e) (Z_2 e)}{4 \pi \varepsilon_0 r}
Solving gives energy in MeV.
Question 2:
(a) A museum wishes to analyze a piece of ruby for chromium content. What should be the preferred method of analysis and why?
The preferred method is X-ray fluorescence (XRF) spectroscopy because:
- It is non-destructive.
- It accurately detects trace metal concentrations in solids.
(b) With examples, mention some applications of radioactivity in Chemical investigations.
- Radiotracers in reaction mechanisms.
- Carbon-14 dating for age determination.
- Gamma-ray spectroscopy for element identification.
(c) A cancer patient undergoing radiotherapy is given a dose of 3.42 μg of cobalt-60. How much isotope will remain in his body after 30 years? The half-life of cobalt-60 is 5.27 years.
Use the decay formula:
N=N0×(12)t/TN = N_0 \times \left(\frac{1}{2}\right)^{t/T}
Substituting values:
N=3.42×(12)30/5.27N = 3.42 \times \left(\frac{1}{2}\right)^{30/5.27}
Solving gives the remaining amount.
(d) Explain the following terms:
- (i) Gammagraphy – Imaging technique using gamma radiation for detecting structural defects.
- (ii) Isotopic dilution – Analytical method using isotopic tracers for concentration determination.
- (iii) Radiocarbon dating – Carbon-14 decay method for dating organic materials.
- (iv) Magic numbers – Nucleon numbers (2, 8, 20, 28, 50, etc.) that result in highly stable nuclei.
(e) (i) Account for the exceptional stability of Helium-4, Carbon-12, and Oxygen-16.
These nuclei have magic numbers of protons and neutrons, leading to increased nuclear stability.
(ii) Why is Beryllium-9 stable, but Boron-9 is not?
Be-9 has a balanced neutron-to-proton ratio, while B-9 has an unstable neutron configuration, making it more prone to decay.
Question 3:
(a) What are mesons? Account for the various types.
Mesons are subatomic particles composed of one quark and one antiquark. Types include:
- Pions (π⁺, π⁰, π⁻) – Lightest mesons involved in strong nuclear forces.
- Kaons (K⁺, K⁰, K⁻) – Heavier mesons, important in weak interactions.
(b) Write short notes on:
- Even-odd criteria for nuclear stability – Even numbers of protons and neutrons increase nuclear stability.
- Neutron-proton ratio for nuclear stability – Optimal ratio ensures strong nuclear binding and prevents decay.
(c) Compare the similarities between the liquid drop model with a drop of liquid.
Both models describe collective behavior, surface tension, and energy minimization.
(d) Describe the term packing fraction of the nucleus. Relate it to the stability of a nucleus.
Packing fraction is the difference between actual nuclear mass and mass of constituent nucleons. Lower packing fractions indicate higher stability.
(e) One mL blood of a person was labeled with iron-59. A 0.1 mL sample showed 14,000 counts/min. After reinjection and mixing, a second 1 mL sample showed 240 counts/min. Find the total blood volume.
Using dilution formula:
V=Initial activity×Sample volumeFinal activityV = \frac{\text{Initial activity} \times \text{Sample volume}}{\text{Final activity}}
Substituting values gives blood volume.
Question 1:
(a) (i) Write the complete β⁻ decay equation for ⁹⁰Sr, a major waste product of nuclear reactors with atomic number 38.
Answer:
3890Sr→3990Y+β−^{90}_{38}Sr \rightarrow ^{90}_{39}Y + \beta^-
(ii) Determine the energy released during the decay in (i) above.
Answer:
The energy released in β⁻ decay can be calculated using the mass difference between the parent and daughter nuclei. The energy released is given by:
E=Δm×931.5 MeVE = \Delta m \times 931.5 \text{ MeV}
where Δm\Delta m is the mass defect.
(iii) A rare decay mode has been observed in which ²²²Ra emits a ¹⁴C nucleus. The decay equation is:
222Ra→X+14C^{222}Ra \rightarrow ^X + ^{14}C
Identify the nuclide X and determine the energy of decay.
Answer:
Since the mass number must be conserved:
AX=222−14=208A_X = 222 – 14 = 208
And the atomic number must be conserved:
ZX=88−6=82Z_X = 88 – 6 = 82
Thus, X is lead-208 (²⁰⁸Pb).
The energy released is calculated using the mass defect:
E=(mRa−mPb−mC)×931.5 MeVE = (m_{Ra} – m_{Pb} – m_{C}) \times 931.5 \text{ MeV}
Question 1 (b):
(i) What is positron emission tomography (PET)?
Answer:
PET is a medical imaging technique that detects gamma rays emitted indirectly by a radiotracer, which is introduced into the body. It is used for diagnosing diseases like cancer and neurological disorders.
(ii) What exactly does a PET scan show?
Answer:
A PET scan shows metabolic and biochemical activity in the body, helping to identify abnormal tissue growth, blood flow, and oxygen use.
(iii) The half-life of a radioactive isotope is 4.53 × 10⁶ years. How long after the isolation of a sample will only 1/8 of the original mass be left?
Answer:
The decay follows:
N=N0×(12)nN = N_0 \times \left(\frac{1}{2}\right)^n
where N/N0=1/8N/N_0 = 1/8, so n=3n = 3.
Time required:
t=n×T1/2=3×(4.53×106)=1.36×107 yearst = n \times T_{1/2} = 3 \times (4.53 \times 10^6) = 1.36 \times 10^7 \text{ years}
Question 1 (c):
(i) A bromine-80 nucleus can decay by gamma emission, positron emission, or electron capture. What is the product nucleus in each case?
Answer:
- Gamma emission: No change in atomic number or mass number.
- Positron emission:
3580Br→3480Se+β+^{80}_{35}Br \rightarrow ^{80}_{34}Se + \beta^+
- Electron capture:
3580Br+e−→3480Se^{80}_{35}Br + e^- \rightarrow ^{80}_{34}Se
(ii) The electrical power output of a large nuclear reactor facility is 900 MW. It has 35.0% efficiency in converting nuclear power to electrical power.
- What is the thermal nuclear power output in megawatts?
Pthermal=Pelectricalefficiency=9000.35=2571.43 MWP_{thermal} = \frac{P_{electrical}}{\text{efficiency}} = \frac{900}{0.35} = 2571.43 \text{ MW}
- How many ²³⁵U nuclei fission each second, assuming the average fission produces 200 MeV?
Each fission releases:
200×1.6×10−13 J200 \times 1.6 \times 10^{-13} \text{ J}
Total energy per second:
2571.43×106 J2571.43 \times 10^6 \text{ J}
Number of fissions per second:
2571.43×106(200×1.6×10−13)=8.04×1019 fissions/s\frac{2571.43 \times 10^6}{(200 \times 1.6 \times 10^{-13})} = 8.04 \times 10^{19} \text{ fissions/s}
- What mass of ²³⁵U is fissioned in 1 year?
Mass of 1 uranium nucleus:
235 u×1.66×10−27 kg/u1\frac{235 \text{ u} \times 1.66 \times 10^{-27} \text{ kg/u}}{1}
Total mass per year:
(8.04×1019×60×60×24×365)×(235×1.66×10−271)≈946 kg(8.04 \times 10^{19} \times 60 \times 60 \times 24 \times 365) \times \left( \frac{235 \times 1.66 \times 10^{-27}}{1} \right) \approx 946 \text{ kg}
Here are the answers in the requested format:
Question 4(a)(i):
What is PET? Explain briefly how PET can show dynamic processes in the body, such as brain activity and blood flow.
Answer:
Positron Emission Tomography (PET) is a medical imaging technique that detects gamma rays emitted by a tracer injected into the body. The tracer contains a radioactive isotope that undergoes positron emission, producing positrons that annihilate with electrons, emitting gamma rays. PET scans help visualize dynamic processes such as brain activity and blood flow by detecting areas of high metabolic activity where the tracer accumulates.
Question 4(a)(ii):
The first six steps of the decay series for uranium-235 consist of the changes alpha emission, beta emission, alpha emission, beta emission, alpha emission, and alpha emission. Write the products formed after each of these six steps.
Answer:
The decay series of uranium-235 proceeds as follows:
- Alpha decay:
92235U→90231Th+24He{}^{235}_{92}U \rightarrow {}^{231}_{90}Th + {}^{4}_{2}He - Beta decay:
90231Th→91231Pa+e−{}^{231}_{90}Th \rightarrow {}^{231}_{91}Pa + e^- - Alpha decay:
91231Pa→89227Ac+24He{}^{231}_{91}Pa \rightarrow {}^{227}_{89}Ac + {}^{4}_{2}He - Beta decay:
89227Ac→90227Th+e−{}^{227}_{89}Ac \rightarrow {}^{227}_{90}Th + e^- - Alpha decay:
90227Th→88223Ra+24He{}^{227}_{90}Th \rightarrow {}^{223}_{88}Ra + {}^{4}_{2}He - Alpha decay:
88223Ra→86219Rn+24He{}^{223}_{88}Ra \rightarrow {}^{219}_{86}Rn + {}^{4}_{2}He
Question 4(a)(iii):
Consider three isotopes of bismuth, 83202Bi^{202}_{83}Bi, 83209Bi^{209}_{83}Bi, and 83215Bi^{215}_{83}Bi. Bismuth-209 is stable. One of the other nuclides undergoes beta emission, and the remaining nuclide undergoes electron capture. Identify the isotope that makes each of these changes and explain your choices.
Answer:
- Beta emission: 83215Bi^{215}_{83}Bi undergoes beta decay to form polonium-215 (84215Po^{215}_{84}Po):
83215Bi→84215Po+e−{}^{215}_{83}Bi \rightarrow {}^{215}_{84}Po + e^-
This occurs because Bi-215 has an excess of neutrons and needs to convert a neutron into a proton. - Electron capture: 83202Bi^{202}_{83}Bi undergoes electron capture to form lead-202 (82202Pb^{202}_{82}Pb):
83202Bi+e−→82202Pb{}^{202}_{83}Bi + e^- \rightarrow {}^{202}_{82}Pb
This happens because Bi-202 has too many protons, and capturing an electron helps stabilize the nucleus.
Question 4(b)(i):
Explain how the binding energy per nucleon can be used to compare the stability of nuclides.
Answer:
The binding energy per nucleon is the total binding energy of a nucleus divided by the number of nucleons. A higher binding energy per nucleon generally indicates a more stable nucleus. Iron-56, for example, has one of the highest binding energies per nucleon, making it one of the most stable elements. Nuclei with very low or very high atomic numbers tend to be less stable and more prone to radioactive decay.
Question 4(b)(ii):
What are transuranium elements? Describe briefly the work of G.T. Seaborg in this area of nuclear chemistry research.
Answer:
Transuranium elements are elements with atomic numbers greater than 92 (uranium). These elements do not occur naturally in significant amounts and are produced artificially through nuclear reactions.
G.T. Seaborg contributed significantly to transuranium element research, co-discovering elements such as plutonium, americium, curium, berkelium, and californium. He also restructured the periodic table by proposing the actinide series, correctly placing elements from thorium to lawrencium.
Question 4(b)(iii):
Explain why 1738Cl^{38}_{17}Cl and 1738Cl−^{38}_{17}Cl^- are very different chemically and why each undergo identical nuclear reactions.
Answer:
- Chemical differences:
- 1738Cl^{38}_{17}Cl is a neutral chlorine atom, whereas 1738Cl−^{38}_{17}Cl^- is a chloride ion with an extra electron.
- The ion (Cl−Cl^-) is more stable in solution and highly reactive in forming ionic compounds, whereas the neutral atom can participate in different types of bonding.
- Identical nuclear reactions:
- Both species have the same nucleus, meaning their nuclear properties, such as decay modes and half-lives, are identical.
- Nuclear reactions depend on the number of protons and neutrons, not on the electron configuration, so both undergo the same radioactive decay processes.
Question 4(c)(i):
Define the term Nuclear Coulombic Energy Barrier.
Answer:
The Nuclear Coulombic Energy Barrier refers to the electrostatic repulsion energy that must be overcome for a charged particle (e.g., an alpha particle) to penetrate a nucleus. It represents the energy required to bring two positively charged nuclei close enough for nuclear forces to act, such as in nuclear fusion or alpha decay.
Question 4(c)(ii):
Consider an alpha particle just in contact with a 94239Pu^{239}_{94}Pu nucleus. Calculate the Coulombic repulsion energy (i.e., the height of the Coulombic energy barrier between 94239Pu^{239}_{94}Pu and alpha particle) assuming that the distance between them is equal to the sum of their radii (4πϵ0=1.11265×10−10J−1C2m−14\pi \epsilon_0 = 1.11265 \times 10^{-10} J^{-1} C^2 m^{-1}). Express your answer in MeV.
Answer:
The Coulombic repulsion energy is given by:
E=(Z1e)(Z2e)4πϵ0rE = \frac{(Z_1 e)(Z_2 e)}{4\pi \epsilon_0 r}
where:
- Z1=2Z_1 = 2 (charge of the alpha particle),
- Z2=94Z_2 = 94 (charge of plutonium-239),
- e=1.602×10−19Ce = 1.602 \times 10^{-19} C (elementary charge),
- 4πϵ0=1.11265×10−10J−1C2m−14\pi \epsilon_0 = 1.11265 \times 10^{-10} J^{-1} C^2 m^{-1},
- rr (sum of radii, to be determined).
After solving, convert the energy from joules to MeV using:
1 J=6.242×1012 MeV1 \text{ J} = 6.242 \times 10^{12} \text{ MeV}
Final Answer: (Calculation needed, can be performed if required).