Class 12 - Unit # 27 : Nuclear Physics - Solved Numericals

1. Radioactive Decay Half-Life Calculation

In 9.0 days the number of radioactive nuclei decreases to one-eighth the number present initially. What is the half-life (in days) of the material?

Data:

• Total Time elapsed (t) = 9.0 days
• Remaining Fraction of Nuclei (N / N₀) = 1/8
• Half-Life (T_1/2) = ?

Solution:

According to the Law of Radioactive Decay: N / N₀ = (1/2)ⁿ, where n is the number of half-lives.

Given fraction: 1/8 = (1/2)³

Equating powers: n = 3

Since n = t / T_1/2, we rearrange to find: T_1/2 = t / n

Calculating the value: T_1/2 = 9.0 / 3 = 3.0 days

Result: The half-life (in days) of the material is 3 days.

2. Decay Constant of Phosphorus Isotope

The phosphorus isotope (P-32) has a half-life of 14.28 days. What is its decay constant in units of s⁻¹?

Data:

• Half-Life (T_1/2) = 14.28 days
• Decay Constant (λ) = ?

Solution:

Convert half-life from days into seconds: T_1/2 = 14.28 × 24 × 60 × 60 s = 1,233,792 s

The relationship between half-life and decay constant is: λ = 0.693 / T_1/2

Substitute the value: λ = 0.693 / 1,233,792

Compute final value: λ ≈ 5.617 × 10⁻⁷ s⁻¹

Result: The decay constant of ³²P is 5.617 × 10⁻⁷ s⁻¹.

3. Binding Energy of Lithium-7

Find the binding energy (in MeV) for lithium (⁷₃Li) with an atomic mass equal to 7.016003 u.

Data:

• Atomic Mass of Lithium-7 (M) = 7.016003 u
• Atomic Number (Z) = 3 (Protons)
• Mass Number (A) = 7 ⇒ Neutron Number (N) = A - Z = 4
• Mass of Proton (m_p) = 1.007276 u
• Mass of Neutron (m_n) = 1.008665 u
• Binding Energy (E_b) = ?

Solution:

First, calculate the mass defect (Δm): Δm = [Z(m_p) + N(m_n)] - M

Δm = [3(1.007276) + 4(1.008665)] - 7.016003

Δm = [3.021828 + 4.034660] - 7.016003

Δm = 7.056488 - 7.016003 = 0.040485 u

Since 1 u = 931.5 MeV, binding energy is: E_b = Δm × 931.5 MeV

E_b = 0.040485 × 931.5 ≈ 37.71 MeV

Result: The binding energy (in MeV) for lithium ⁷₃Li is 37.71 MeV.

4. Mass Defect from Binding Energy

The binding energy of a nucleus is 225.0 MeV. What is the mass defect of the nucleus in atomic mass units?

Data:

• Binding Energy (E_b) = 225.0 MeV
• Mass Defect (Δm) = ?

Solution:

The relation between binding energy and mass defect is: E_b = Δm × 931.5 MeV

Rearranging to find mass defect: Δm = E_b / 931.5

Substitute the value: Δm = 225.0 / 931.5

Compute the value: Δm ≈ 0.2415 u

Result: The mass defect of the nucleus in atomic mass units is 0.2415 u.

5. Total Disassembly Energy of a Copper Penny

A copper penny has a mass of 3.0 g. Determine the energy (in MeV) that would be required to break all the copper nuclei into their constituent protons and neutrons. Assume all copper nuclei are ⁶³₂₉Cu (atomic mass = 62.939598 u).

Data:

• Mass of Copper Sample (m) = 3.0 g
• Molar Mass of Copper-63 (M) = 63 g/mol
• Atomic Number (Z) = 29, Neutron Number (N) = 63 - 29 = 34
• Atomic Mass of ⁶³Cu = 62.939598 u
• Avogadro's Number (N_A) = 6.022 × 10²³ atoms/mol
• Total Disassembly Energy (E_total) = ?

Solution:

Calculate mass defect (Δm) for a single copper nucleus: Δm = [29(1.007276) + 34(1.008665)] - 62.939598

Δm = [29.211004 + 34.294610] - 62.939598 = 63.505614 - 62.939598 = 0.566016 u

Calculate binding energy of one copper atom: E_b = 0.566016 × 931.5 MeV ≈ 527.24 MeV

Find total number of nuclei (N_atoms) in 3.0 g of copper: N_atoms = (m / M) × N_A

N_atoms = (3.0 / 63) × 6.022 × 10²³ ≈ 2.868 × 10²² nuclei

Calculate total energy required: E_total = N_atoms × E_b

E_total = 2.868 × 10²² × 527.24 MeV ≈ 1.512 × 10²⁵ MeV

Result: The total energy required to break all the copper nuclei into constituents is 1.512 × 10²⁵ MeV.

6. Beta Plus (β⁺) Decay Processes

Write the Beta+ decay process for each of the following nuclei with their proper chemical symbols including Z and A for each daughter nucleus: (a) ¹¹₆C (b) ¹⁵₈O

Solution:

During Beta+ (β⁺) decay, a proton transforms into a neutron, emitting a positron (e⁺) and an electron neutrino (ν_e). The atomic number decreases by 1, while the mass number stays the same.

(a) Decay process for Carbon-11 (¹¹₆C):

¹¹₆C → ¹¹₅B + ₊₁⁰e + ν_e (Daughter nucleus is Boron-11)

(b) Decay process for Oxygen-15 (¹⁵₈O):

¹⁵₈O → ¹⁵₇N + ₊₁⁰e + ν_e (Daughter nucleus is Nitrogen-15)

7. Activity of a Cobalt-60 Radiation Therapy Source

A device used in radiation therapy for cancer contains 0.50 g of cobalt-60 (59.933819 u). The half-life of ⁶⁰Co is 5.27 years. Determine the activity of the radioactive material.

Data:

• Mass of Cobalt sample (m) = 0.50 g
• Molar Mass / Atomic Mass of ⁶⁰Co ≈ 60 g/mol
• Half-Life (T_1/2) = 5.27 years = 5.27 × 365 × 24 × 3600 s ≈ 1.662 × 10⁸ s
• Avogadro's Number (N_A) = 6.022 × 10²³ atoms/mol
• Activity of Sample (A) = ?

Solution:

First, find the total number of radioactive nuclei (N): N = (m / M) × N_A

N = (0.50 / 60) × 6.022 × 10²³ ≈ 5.018 × 10²¹ nuclei

Next, calculate the decay constant (λ): λ = 0.693 / T_1/2

λ = 0.693 / (1.662 × 10⁸) ≈ 4.169 × 10⁻⁹ s⁻¹

Calculate the radioactive activity (A): A = λ × N

A = 4.169 × 10⁻⁹ × 5.018 × 10²¹ ≈ 2.092 × 10¹³ Bq (or disintegrations per second)

In Curies (Optional): A = (2.092 × 10¹³) / (3.7 × 10¹⁰) ≈ 565.4 Ci

Result: The activity of the radioactive material is 2.092 × 10¹³ Bq (or 565.4 Ci).