Let's break down part (a) and (b) into two clear steps as requested: --- ## **Step 1: Calculate the Activity (R) Due to ⁴⁰K** ### **Given Data** - **Person's mass (m):** 76 kg - **Fraction of body mass that is potassium:** .20% = .002 - **Fraction of potassium that is ⁴⁰K:** .0117% = .000117 - **Molar mass of potassium:** ~39.1 g/mol - **Half-life (T₁/₂) of ⁴⁰K:** 1.28 × 10⁹ years - **Decay energy per reaction:** 1.3 MeV (for next step) - **Avogadro's Number (Nₐ):** 6.022 × 10²³ mol⁻¹ ### **Find the Number of ⁴⁰K Nuclei (N)** - Mass of potassium in body: \( m_{\text{K}} = 76\,\text{kg} \times .002 = .152\,\text{kg} = 152\,\text{g} \) - Moles of potassium: \( n_{\text{K}} = \frac{152\,\text{g}}{39.1\,\text{g/mol}} = 3.89\,\text{mol} \) - Number of potassium atoms: \( N_{\text{K}} = 3.89\,\text{mol} \times 6.022 \times 10^{23}\,\text{atoms/mol} = 2.34 \times 10^{24}\,\text{atoms} \) - Number of ⁴⁰K nuclei: \( N_{40K} = 2.34 \times 10^{24} \times .000117 = 2.74 \times 10^{20} \) ### **Decay Constant (λ)** \[ λ = \frac{\ln(2)}{T_{1/2}} \] - \(T_{1/2} = 1.28 \times 10^9\,\text{years} = 1.28 \times 10^9 \times 3.156 \times 10^7\,\text{s} = 4.04 \times 10^{16}\,\text{s}\) - \(λ = \frac{.693}{4.04 \times 10^{16}\,\text{s}} = 1.71 \times 10^{-17}\,\text{s}^{-1}\) ### **Calculate Activity (R)** \[ R = N \lambda = 2.74 \times 10^{20} \times 1.71 \times 10^{-17} = 4.68 \times 10^3\,\text{Bq} \] **Convert to kBq:** \( 4.68 \times 10^3\,\text{Bq} = 4.68\,\text{kBq} \) --- ### **End of Step 1 Explanation** We found the activity of ⁴⁰K in a 76 kg person is **4.68 kBq**. --- ## **Step 2: Calculate the Effective Dose in mSv/year** ### **Energy per decay:** - \( E_{\text{decay}} = 1.3\,\text{MeV} = 1.3 \times 1.602 \times 10^{-13}\,\text{J} = 2.08 \times 10^{-13}\,\text{J} \) ### **Total energy per year:** - Decays per year: \( 4.68 \times 10^3\,\text{decays/s} \times 3.156 \times 10^7\,\text{s/year} = 1.48 \times 10^{11}\,\text{decays/year} \) - Total energy: \( 1.48 \times 10^{11} \times 2.08 \times 10^{-13}\,\text{J} = .0308\,\text{J/year} \) ### **Dose (Gy):** \[ \text{Dose} = \frac{\text{Energy}}{\text{mass}} = \frac{.0308\,\text{J}}{76\,\text{kg}} = 4.05 \times 10^{-4}\,\text{Gy/year} \] - 1 Gy = 1 J/kg ### **Effective Dose (mSv):** - For beta and gamma, quality factor \( Q = 1 \) \[ \text{Effective Dose} = 4.05 \times 10^{-4}\,\text{Gy/year} \times 100 = .405\,\text{mSv/year} \] --- ### **End of Step 2 Explanation** The effective dose due to ⁴⁰K is **.405 mSv/year**. --- ## **Final Answers** - **(a) Activity inside the body due to ⁴⁰K:** \(\boxed{4.68\,\text{kBq}}\) - **(b) Effective dose:** \(\boxed{.405\,\text{mSv/year}}\)

# Internal Radiation Dose Calculation We will calculate the activity due to potassium-40 (⁴⁰K) and the effective dose in two steps, providing explanations after each step. --- ## **Step 1: Calculate the Activity (R) Due to ⁴⁰K** ### **Given Data** - **Person's mass (m):** 76 kg - **Fraction of body mass that is potassium:** 0.20% = 0.002 - **Fraction of potassium that is ⁴⁰K:** 0.0117% = 0.000117 - **Molar mass of potassium:** ~39.1 g/mol - **Half-life (T₁/₂) of ⁴⁰K:** 1.28 × 10⁹ years - **Decay energy per reaction:** 1.3 MeV - **Avogadro's Number (Nₐ):** 6.022 × 10²³ mol⁻¹ ### **Calculations** 1. **Mass of potassium in the body:** \[ m_{\text{K}} = 76\,\text{kg} \times 0.002 = 0.152\,\text{kg} = 152\,\text{g} \] 2. **Moles of potassium:** \[ n_{\text{K}} = \frac{152\,\text{g}}{39.1\,\text{g/mol}} = 3.89\,\text{mol} \] 3. **Number of potassium atoms:** \[ N_{\text{K}} = 3.89\,\text{mol} \times 6.022 \times 10^{23}\,\text{atoms/mol} = 2.34 \times 10^{24}\,\text{atoms} \] 4. **Number of ⁴⁰K nuclei:** \[ N_{40K} = 2.34 \times 10^{24} \times 0.000117 = 2.74 \times 10^{20} \] 5. **Decay Constant (λ):** \[ λ = \frac{\ln(2)}{T_{1/2}} = \frac{0.693}{1.28 \times 10^9\,\text{years} \times 3.156 \times 10^7\,\text{s/year}} = 1.71 \times 10^{-17}\,\text{s}^{-1} \] 6. **Calculate Activity (R):** \[ R = N_{40K} \lambda = 2.74 \times 10^{20} \times 1.71 \times 10^{-17} = 4.68 \times 10^3\,\text{Bq} \] **Convert to kBq:** \[ 4.68 \times 10^3\,\text{Bq} = 4.68\,\text{kBq} \] ### **End of Step 1 Explanation** We have calculated the activity of ⁴⁰K in a 76 kg person to be **4.68 kBq**. --- ## **Step 2: Calculate the Effective Dose in mSv/year** ### **Given Data** - **Energy per decay:** \[ E_{\text{decay}} = 1.3\,\text{MeV} = 1.3 \times 1.602 \times 10^{-13}\,\text{J} = 2.08 \times 10^{-13}\,\text{J} \] ### **Calculations** 1. **Total decays per year:** \[ \text{Decays per year} = 4.68 \times 10^3\,\text{decays/s} \times 3.156 \times 10^7\,\text{s/year} = 1.48 \times 10^{11}\,\text{decays/year} \] 2. **Total energy from decays per year:** \[ \text{Total energy} = 1.48 \times 10^{11} \times 2.08 \times 10^{-13}\,\text{J} = 0.0308\,\text{J/year} \] 3. **Dose (Gy):** \[ \text{Dose} = \frac{\text{Energy}}{\text{mass}} = \frac{0.0308\,\text{J}}{76\,\text{kg}} = 4.05 \times 10^{-4}\,\text{Gy/year} \] 4. **Effective Dose (mSv):** \[ \text{Effective Dose} = 4.05 \times 10^{-4}\,\text{Gy/year} \times 1000 = 0.405\,\text{mSv/year} \] ### **End of Step 2 Explanation** The effective dose due to ⁴⁰K is **0.405 mSv/year**. --- ## **Final Answers** - **(a) Activity inside the body due to ⁴⁰K:** \(\boxed{4.68\,\text{kBq}}\) - **(b) Effective dose:** \(\boxed{0.405\,\text{mSv/year}}\)