Exercises
Ex1. Nuclear energy and number of nucleus
Part A : A cobalt nucleus 6027Co, in an excited state, emits γ radiation with energy 1.33 MeV when returning to the ground state.
Its mass in the ground state is 59.93382 u. Find its mass in the excited state.
Help : Enucleus = mc² + Ec,nucleus
Part B : Calculating Number of Nuclei
A sample contains 5 mg of radium-226 (22688Ra) where each nucleus has a mass of 225.9770 u.
Given:
• Sample mass = 5 mg = 5×10-6 kg
• Mass per nucleus = 225.9770 u
• 1 u = 1.66×10-27 kg
Calculate: The number of nuclei in this sample.
Part A : A cobalt nucleus 6027Co, in an excited state, emits γ radiation with energy 1.33 MeV when returning to the ground state.
Its mass in the ground state is 59.93382 u. Find its mass in the excited state.
Help : Enucleus = mc² + Ec,nucleus
Part B : Calculating Number of Nuclei
A sample contains 5 mg of radium-226 (22688Ra) where each nucleus has a mass of 225.9770 u.
Given:
• Sample mass = 5 mg = 5×10-6 kg
• Mass per nucleus = 225.9770 u
• 1 u = 1.66×10-27 kg
Calculate: The number of nuclei in this sample.
Part C. Gamma Radiation (γ)
The reaction: AZ+1Y* → AZ+1Y + γ
Determine the wavelength of a photon emitted by a magnesium-24 nucleus (²⁴₁₂Mg) when it transitions from energy level E₃ = 5.22 MeV to E₂ = 4.12 MeV.
The reaction: AZ+1Y* → AZ+1Y + γ
Determine the wavelength of a photon emitted by a magnesium-24 nucleus (²⁴₁₂Mg) when it transitions from energy level E₃ = 5.22 MeV to E₂ = 4.12 MeV.
🔸 Given:
- Planck’s constant: h = 6.63 × 10⁻³⁴ J·s
- Speed of light: c = 3 × 10⁸ m/s
- Planck’s constant: h = 6.63 × 10⁻³⁴ J·s
- Speed of light: c = 3 × 10⁸ m/s
❓ Question:
- Calculate the wavelength of the emitted gamma photon in meters (m).
- Calculate the wavelength of the emitted gamma photon in meters (m).
Ex2. Nuclear reactions//
Soddy's Laws
Conservation of mass number: ∑Abefore = ∑Aafter
Conservation of charge number: ∑Zbefore = ∑Zafter
Part A : Complete the nuclear reactions
a. 4520Ca → ... + 0-1e + 00ν
b. 5829Cu* → ... + γ
c. 4624Cr → 4623V + ...
d. 23494Pu → ... + 42He
e. 23993Np → 23994Pu + ... + ...
f. 158O → ... + 01e + ν
a. 4520Ca → ... + 0-1e + 00ν
b. 5829Cu* → ... + γ
c. 4624Cr → 4623V + ...
d. 23494Pu → ... + 42He
e. 23993Np → 23994Pu + ... + ...
f. 158O → ... + 01e + ν
Part B : Cobalt 6027Co is β- radioactive
The daughter nucleus is Nickel 6028Ni.
Determine Z and A.
The daughter nucleus is Nickel 6028Ni.
Determine Z and A.
Part C : Decay of uranium 238
The balance equation for the decay of uranium 238 that leads to lead 206 is:
23892U → 20682Pb + xβ- + yα
Determine, specifying the laws used, the values of x and y.
The balance equation for the decay of uranium 238 that leads to lead 206 is:
23892U → 20682Pb + xβ- + yα
Determine, specifying the laws used, the values of x and y.
Part D : Transformation of uranium 235
A uranium nucleus 23592U transforms, after a series of α and β- decays, into 20782Pb.
Determine the numbers x and y of these α and β- decays respectively.
A uranium nucleus 23592U transforms, after a series of α and β- decays, into 20782Pb.
Determine the numbers x and y of these α and β- decays respectively.
Ex3. α decay of Radium ; 22688Ra
Generally: We will applicate ⤵
Elib = Ec(daughter) - Ec(parent) + ∑Ec(particles) + Eγ
Radium 22688Ra decays spontaneously by emitting an α particle:
22688Ra → 22286Rn + 42He + γ
Data:
m(22688Ra) = 225.9770 u
m(22286Rn) = 221.9702 u
m(42He) = 4.0015 u
1u = 931.5 MeV/c2 = 1.6605×10-27 kg
1.a) Calculate, in MeV, the energy released by this decay.
2.b) If the energy of the obtained γ radiation is 0.6 MeV, determine the kinetic energy acquired by the α particle. Deduce its exit velocity.
Generally: We will applicate ⤵
Elib = Ec(daughter) - Ec(parent) + ∑Ec(particles) + Eγ
Radium 22688Ra decays spontaneously by emitting an α particle:
22688Ra → 22286Rn + 42He + γ
Data:
m(22688Ra) = 225.9770 u
m(22286Rn) = 221.9702 u
m(42He) = 4.0015 u
1u = 931.5 MeV/c2 = 1.6605×10-27 kg
1.a) Calculate, in MeV, the energy released by this decay.
2.b) If the energy of the obtained γ radiation is 0.6 MeV, determine the kinetic energy acquired by the α particle. Deduce its exit velocity.
Ex4. Conservation of Total Energy
Consider the alpha decay of Radium 226 (22688Ra) into Radon 222 (22286Rn) and an alpha particle (42He). The atomic masses are:
- m(22688Ra) = 226.0254 u
- m(22286Rn) = 222.0176 u
- m(42He) = 4.0026 u
- 1 u = 931.5 MeV/c²
- Calculate the mass defect (Δm) of this decay in atomic mass units (u) and in MeV/c².
- Calculate the total energy released (Q) during this decay in MeV.
Ex5. Formation of carbon 14
In the upper atmosphere, nitrogen 147N transforms, under the impact of a neutron, into 146C, an isotope of carbon 126C.
1. The nuclides 126C and 146C are isotopes. Give the components of each of these two nuclides and explain why they are called carbon isotopes?
2. Write the equation for the formation reaction of 146C.
3. Is this nuclear reaction spontaneous? Justify.
4. Identify the emitted particle.
In the upper atmosphere, nitrogen 147N transforms, under the impact of a neutron, into 146C, an isotope of carbon 126C.
1. The nuclides 126C and 146C are isotopes. Give the components of each of these two nuclides and explain why they are called carbon isotopes?
2. Write the equation for the formation reaction of 146C.
3. Is this nuclear reaction spontaneous? Justify.
4. Identify the emitted particle.
Ex6. Activity of a Sample
A sample contains 1.0 × 1015 nuclei of a radioactive isotope with a half-life of 12 hours.
- Calculate the decay constant (λ) of this isotope in s-1.
- Calculate the initial activity of the sample in Becquerels (Bq).
Ex7. Relationship between Activity and Number of Nuclei
The activity of a Sodium 24 (2411Na) sample is 2.5 MBq. The half-life of Sodium 24 is 15 hours.
- Calculate the decay constant of Sodium 24.
- Determine the number of Sodium 24 nuclei present in the sample.
Ex8. α DECAY
AZX → A-4Z-2Y + 42He + γ
Polonium decay and γ radiation
Data: h = 6.63×10-34 J·s ; c = 3×108 m/s.
Polonium 21084Po is an α emitter. The produced daughter nucleus is a lead nucleus 20682Pb in an excited state.
1. What causes the emission of γ radiation?
2. Specify its nature.
3. Calculate the energy of a γ photon with wavelength λ = 1.35×10-12 m.
AZX → A-4Z-2Y + 42He + γ
Polonium decay and γ radiation
Data: h = 6.63×10-34 J·s ; c = 3×108 m/s.
Polonium 21084Po is an α emitter. The produced daughter nucleus is a lead nucleus 20682Pb in an excited state.
1. What causes the emission of γ radiation?
2. Specify its nature.
3. Calculate the energy of a γ photon with wavelength λ = 1.35×10-12 m.
Ex9. Activity of a Radioactive Sample (Carbon-14)
The activity of a substance containing carbon-14 is determined using a β⁻ particle counter. A wood sample containing 0.05 g of carbon-14 is exposed to this detector. The radioactive half-life of carbon-14 is T = 5570 years.
The activity of a substance containing carbon-14 is determined using a β⁻ particle counter. A wood sample containing 0.05 g of carbon-14 is exposed to this detector. The radioactive half-life of carbon-14 is T = 5570 years.
🔸 Given Data:
- Mass of a proton: mp = 1.00728 u
- Mass of a ¹⁴C nucleus: 14.0065 u
- Mass of a neutron: mn = 1.00866 u
- Mass of a ¹⁴N nucleus: 14.0031 u
- Molar mass of ¹⁴C: 14 g·mol⁻¹
- Avogadro’s number: NA = 6.02 × 10²³ mol⁻¹
- 1 u = 931.5 MeV/c²
- Mass of a proton: mp = 1.00728 u
- Mass of a ¹⁴C nucleus: 14.0065 u
- Mass of a neutron: mn = 1.00866 u
- Mass of a ¹⁴N nucleus: 14.0031 u
- Molar mass of ¹⁴C: 14 g·mol⁻¹
- Avogadro’s number: NA = 6.02 × 10²³ mol⁻¹
- 1 u = 931.5 MeV/c²
❓ Determine:
- The decay constant λ of carbon-14
- The number of carbon-14 nuclei in the sample (0.05 g)
- The activity of the sample at the given time
- The decay constant λ of carbon-14
- The number of carbon-14 nuclei in the sample (0.05 g)
- The activity of the sample at the given time
Ex10. 21283Bi
α particle accompanied by γ radiation
Bismuth 21283Bi is α and γ radioactive. The daughter nucleus is an isotope of thallium Tl.
Data:
mBi = 211.991876 u
mTl = 207.982013 u
mα = 4.0015 u
1. What causes the presence of γ radiation?
2. Write the balance equation (1) for γ de-excitation.
3. Write the balance equation (2) for the bismuth nucleus decay. Deduce the components of the thallium nucleus.
4. Determine the energy released by the nuclear reaction (2).
5. Knowing that the wavelength of γ radiation is 3.78 pm (1 pm = 10-12 m), determine the kinetic energy of α radiation and deduce its velocity.
α particle accompanied by γ radiation
Bismuth 21283Bi is α and γ radioactive. The daughter nucleus is an isotope of thallium Tl.
Data:
mBi = 211.991876 u
mTl = 207.982013 u
mα = 4.0015 u
1. What causes the presence of γ radiation?
2. Write the balance equation (1) for γ de-excitation.
3. Write the balance equation (2) for the bismuth nucleus decay. Deduce the components of the thallium nucleus.
4. Determine the energy released by the nuclear reaction (2).
5. Knowing that the wavelength of γ radiation is 3.78 pm (1 pm = 10-12 m), determine the kinetic energy of α radiation and deduce its velocity.
Ex11. Radioactivity of polonium 21084Po
To study the radioactivity of polonium 21084Po which is an α emitter, we have a sample of polonium 210 containing N0 nuclei at time t0 = 0.
We measure, at successive dates, the number N of remaining nuclei. We calculate the ratio N/N0 and draw up the following table:
t (days) 0 50 100 150 200 250 300 N/N0 1 0,78 0,61 0,47 0,37 0,29 0,22 -ln(N/N0) 0 0,25 1,24
- Reproduce and complete this table by calculating -ln(N/N0) at each date.
2. Plot the curve representing the evolution of f(t) = -ln(N/N0) as a function of time at the scale of 1 cm on the x-axis for 20 days and 1 cm on the y-axis for 0.1.
3.1) Knowing that ln(N/N0) = -λt, determine graphically the value of the radioactive constant λ of polonium 210.
3.2) Deduce the half-life of polonium 210.
To study the radioactivity of polonium 21084Po which is an α emitter, we have a sample of polonium 210 containing N0 nuclei at time t0 = 0.
We measure, at successive dates, the number N of remaining nuclei. We calculate the ratio N/N0 and draw up the following table:
| t (days) | 0 | 50 | 100 | 150 | 200 | 250 | 300 |
|---|---|---|---|---|---|---|---|
| N/N0 | 1 | 0,78 | 0,61 | 0,47 | 0,37 | 0,29 | 0,22 |
| -ln(N/N0) | 0 | 0,25 | 1,24 |
- Reproduce and complete this table by calculating -ln(N/N0) at each date.
2. Plot the curve representing the evolution of f(t) = -ln(N/N0) as a function of time at the scale of 1 cm on the x-axis for 20 days and 1 cm on the y-axis for 0.1.
3.1) Knowing that ln(N/N0) = -λt, determine graphically the value of the radioactive constant λ of polonium 210.
3.2) Deduce the half-life of polonium 210.
Ex12. Decay of Radium 22688Ra
A radium nucleus 22688Ra is radioactive, an α emitter, and can decay into a daughter nucleus, radon Rn.
a) Decay equation
Write the balance equation for this decay.
b) Energy diagram
The figure above shows the energy diagram of radon.
Calculate the wavelength of the radiation emitted during transition 1.
c) Radioactive period
The period of radium is 1620 years.
Calculate the activity of a freshly prepared 0.1 g sample of this element.
Data:
Planck constant: h = 6.63×10-34 J·s
Avogadro's number: NA = 6.022×1023 mol-1
1 year = 365.25 days
A radium nucleus 22688Ra is radioactive, an α emitter, and can decay into a daughter nucleus, radon Rn.
a) Decay equation
Write the balance equation for this decay.
b) Energy diagram
The figure above shows the energy diagram of radon.
Calculate the wavelength of the radiation emitted during transition 1.
c) Radioactive period
The period of radium is 1620 years.
Calculate the activity of a freshly prepared 0.1 g sample of this element.
Data:
Planck constant: h = 6.63×10-34 J·s
Avogadro's number: NA = 6.022×1023 mol-1
1 year = 365.25 days
Ex13. α decay of Radium ; 22688Ra
Generally: We will applicate ⤵
Elib = Ec(daughter) - Ec(parent) + ∑Ec(particles) + Eγ
Radium 22688Ra decays spontaneously by emitting an α particle:
22688Ra → 22286Rn + 42He + γ
Data:
m(22688Ra) = 225.9770 u
m(22286Rn) = 221.9702 u
m(42He) = 4.0015 u
1u = 931.5 MeV/c2 = 1.6605×10-27 kg
1.a) Calculate, in MeV, the energy released by this decay.
2.b) If the energy of the obtained γ radiation is 0.6 MeV, determine the kinetic energy acquired by the α particle. Deduce its exit velocity.
Generally: We will applicate ⤵
Elib = Ec(daughter) - Ec(parent) + ∑Ec(particles) + Eγ
Radium 22688Ra decays spontaneously by emitting an α particle:
22688Ra → 22286Rn + 42He + γ
Data:
m(22688Ra) = 225.9770 u
m(22286Rn) = 221.9702 u
m(42He) = 4.0015 u
1u = 931.5 MeV/c2 = 1.6605×10-27 kg
1.a) Calculate, in MeV, the energy released by this decay.
2.b) If the energy of the obtained γ radiation is 0.6 MeV, determine the kinetic energy acquired by the α particle. Deduce its exit velocity.
Ex14. α decay and kinetic energy
Data:
1u = 1.66 × 10-27 kg = 931.5 MeV/c2
h = 6.63 × 10-34 J·s
1 MeV = 1.6 × 10-13 J
c = 3 × 108 m/s
Polonium 21084Po is an α emitter. The daughter nucleus produced by this decay is a lead nucleus 20682Pb.
This α decay is accompanied by the emission of γ radiation.
This γ radiation is emitted with a vacuum wavelength λ = 1.35 × 10-12 m.
Using the conservation of total energy, determine the kinetic energy of the emitted α particle knowing that the energy released by this decay has the value E = 5.4 MeV.
Data:
1u = 1.66 × 10-27 kg = 931.5 MeV/c2
h = 6.63 × 10-34 J·s
1 MeV = 1.6 × 10-13 J
c = 3 × 108 m/s
Polonium 21084Po is an α emitter. The daughter nucleus produced by this decay is a lead nucleus 20682Pb.
This α decay is accompanied by the emission of γ radiation.
This γ radiation is emitted with a vacuum wavelength λ = 1.35 × 10-12 m.
Using the conservation of total energy, determine the kinetic energy of the emitted α particle knowing that the energy released by this decay has the value E = 5.4 MeV.
Ex15. α particle not accompanied by γ radiation
Data:
1 MeV = 1.6 × 10-13 J
Polonium 21084Po is an α emitter. The daughter nucleus produced by this decay is a lead nucleus 20682Pb.
Knowing that:
• The energy released by this decay has the value E = 5.4 MeV
• The parent nucleus 21084Po is at rest
• The daughter Pb nucleus is born almost motionless and in the ground state
Determine, in J, the kinetic energy of the emitted α particle.
Data:
1 MeV = 1.6 × 10-13 J
Polonium 21084Po is an α emitter. The daughter nucleus produced by this decay is a lead nucleus 20682Pb.
Knowing that:
• The energy released by this decay has the value E = 5.4 MeV
• The parent nucleus 21084Po is at rest
• The daughter Pb nucleus is born almost motionless and in the ground state
Determine, in J, the kinetic energy of the emitted α particle.
Ex16.
Radioactive Half-Life
Cobalt 60 (60Co) has a half-life of 5.27 years.
- What fraction of a Cobalt 60 sample will remain after 10.54 years?
- How long will it take for the activity of a Cobalt 60 sample to decrease by 75%?
Cobalt 60 (60Co) has a half-life of 5.27 years.
- What fraction of a Cobalt 60 sample will remain after 10.54 years?
- How long will it take for the activity of a Cobalt 60 sample to decrease by 75%?
Ex17. Activity of a Sample
A sample contains 1.0 × 1015 nuclei of a radioactive isotope with a half-life of 12 hours.
- Calculate the decay constant (λ) of this isotope in s-1.
- Calculate the initial activity of the sample in Becquerels (Bq).
A sample contains 1.0 × 1015 nuclei of a radioactive isotope with a half-life of 12 hours.
- Calculate the decay constant (λ) of this isotope in s-1.
- Calculate the initial activity of the sample in Becquerels (Bq).
Ex18. Relationship between Activity and Number of Nuclei
The activity of a Sodium 24 (2411Na) sample is 2.5 MBq. The half-life of Sodium 24 is 15 hours.
- Calculate the decay constant of Sodium 24.
- Determine the number of Sodium 24 nuclei present in the sampleCarbon-14 Dating and Specific Activity
The activity of a Sodium 24 (2411Na) sample is 2.5 MBq. The half-life of Sodium 24 is 15 hours.
- Calculate the decay constant of Sodium 24.
- Determine the number of Sodium 24 nuclei present in the sampleCarbon-14 Dating and Specific Activity
An archaeological sample of ancient wood shows a Carbon-14 (146C) activity of 3.9 disintegrations per minute per gram of total carbon. The specific activity of Carbon-14 in living organisms is known to be 15.3 disintegrations per minute per gram of carbon. The half-life of Carbon-14 is 5730 years.
- Determine the age of this wood sample. Detail the steps of your calculation.
- Explain why the activity of Carbon-14 in living organisms is considered constant and the factors that might affect this constancy over very long timescales.
Ex19. Carbon-14 Dating and Specific Activity
An archaeological sample of ancient wood shows a Carbon-14 (146C) activity of 3.9 disintegrations per minute per gram of total carbon. The specific activity of Carbon-14 in living organisms is known to be 15.3 disintegrations per minute per gram of carbon. The half-life of Carbon-14 is 5730 years.
- Determine the age of this wood sample. Detail the steps of your calculation.
- Explain why the activity of Carbon-14 in living organisms is considered constant and the factors that might affect this constancy over very long timescales.
Ex20. Radioactive Half-Life of Polonium 21084Po
The figure below shows the curve representing the variation, as a function of time t, in the number N of nuclei present in a sample of a radioactive substance.
This number is N₀ at time t₀ = 0.
The same figure also shows the tangent to this curve at time t₁ = 263 days.
The figure below shows the curve representing the variation, as a function of time t, in the number N of nuclei present in a sample of a radioactive substance.
This number is N₀ at time t₀ = 0.
The same figure also shows the tangent to this curve at time t₁ = 263 days.
1) Write the expression of N as a function of t and specify the meaning of each term.
2) The radioactive activity of the sample is given by: A = - dN/dt
2.1) Define the radioactive activity A.
2.2) With reference to the figure, determine the value of A at date t₁ = 263 days.
3) Deduce the value of the decay constant and the value of the radioactive half-life (half-life) of polonium 210.
Solution de ces exercices للمشتركين فقط... عذراً
Exercices
Supplémentaires
Chap.19 Radioactivité
Pour obtenir ces exercices, vous devez inscrire يلزم الإشتراك |
To access these exercises, you must register يلزم الإشتراك |
No comments:
Post a Comment