Let's carefully analyze and solve each part of this problem step by step. --- ### **Given Data:** - **Radioisotope:** Phosphorus-32 (\( ^{32}_15 P \)) - **Type of decay:** Beta-minus emitter - **Half-life (\( T_{1/2} \))**: 14.3 days - **Average kinetic energy of emitted electrons (\( E_{avg} \))**: \( 7.00 \times 10^2 \) keV = 700 keV - **Initial activity (\( A_0 \))**: 1.31 MBq = \( 1.31 \times 10^6 \) disintegrations/sec - **Time period:** 10 days - **Mass of tissue:** \( 10^2 \) g = 100 g --- ## **Part (a): Number of electrons emitted in 10 days** ### Step 1: Find the decay constant \( \lambda \) The decay constant \( \lambda \) relates to the half-life: \[ \lambda = \frac{\ln 2}{T_{1/2}} \] Convert half-life to seconds: \[ T_{1/2} = 14.3\, \text{days} \times 86400\, \text{sec/day} = 14.3 \times 86400 \approx 1,236,000\, \text{sec} \] Calculate \( \lambda \): \[ \lambda = \frac{\ln 2}{1,236,000} \approx \frac{0.693}{1,236,000} \approx 5.61 \times 10^{-7}\, \text{sec}^{-1} \] --- ### Step 2: Find the activity \( A(t) \) at time \( t = 10 \) days Convert 10 days to seconds: \[ t = 10 \times 86400 = 864,000\, \text{sec} \] The activity at time \( t \): \[ A(t) = A_0 e^{-\lambda t} \] Calculate \( A(t) \): \[ A(10\, \text{days}) = 1.31 \times 10^6 \times e^{-5.61 \times 10^{-7} \times 864,000} \] Calculate the exponent: \[ - \lambda t = -5.61 \times 10^{-7} \times 864,000 \approx -0.485 \] Calculate \( e^{-0.485} \): \[ e^{-0.485} \approx 0.615 \] Thus, \[ A(10\, \text{days}) \approx 1.31 \times 10^6 \times 0.615 \approx 805,000\, \text{disintegrations/sec} \] --- ### Step 3: Find total disintegrations over 10 days Total disintegrations \( N_{dis} \): \[ N_{dis} = \int_0^{T} A(t) dt \] Since activity decreases exponentially: \[ N_{dis} = \frac{A_0}{\lambda} \left( 1 - e^{-\lambda T} \right) \] Alternatively, for a constant activity approximation, since activity decreases but over a relatively short period compared to the decay, using the average activity: \[ A_{avg} = \frac{A_0 + A(T)}{2} \approx \frac{1.31 \times 10^6 + 8.05 \times 10^5}{2} = \frac{2.115 \times 10^6}{2} = 1.0575 \times 10^6\, \text{dis/sec} \] But to be precise, let's use the integral formula: \[ N_{dis} = \frac{A_0}{\lambda} (1 - e^{-\lambda T}) \] Calculate \( 1 - e^{-\lambda T} \): \[ 1 - 0.615 = 0.385 \] Calculate \( N_{dis} \): \[ N_{dis} = \frac{1.31 \times 10^6}{5.61 \times 10^{-7}} \times 0.385 \] \[ N_{dis} \approx 2.336 \times 10^{12} \times 0.385 \approx 9.0 \times 10^{11} \] **Therefore:** \[ \boxed{ \textbf{Number of disintegrations in 10 days} \approx 9.0 \times 10^{11} } \] ### **Step 4: Number of electrons emitted** Each disintegration emits **one beta particle (electron)**, so: \[ \boxed{ \text{Number of electrons emitted} \approx 9.0 \times 10^{11} } \] --- ## **Part (b): Total energy deposited in the body during 10 days** ### Step 1: Total energy per electron Given: \[ E_{avg} = 700\, \text{keV} \] Convert to Joules: \[ 1\, \text{keV} = 1.602 \times 10^{-16}\, \text{J} \] \[ E_{electron} = 700 \times 1.602 \times 10^{-16} \approx 1.121 \times 10^{-13}\, \text{J} \] ### Step 2: Total energy deposited Total energy: \[ E_{total} = N_{dis} \times E_{electron} \] \[ E_{total} = 9.0 \times 10^{11} \times 1.121 \times 10^{-13} \approx 1.009 \times 10^{-1}\, \text{J} \] **Final answer:** \[ \boxed{ \textbf{Total energy deposited} \approx 0.101\, \text{J} } \] --- ## **Part (c): Absorbed dose in tissue** ### Step 1: Define dose \( D \) \[ D = \frac{\text{Energy deposited}}{\text{Mass of tissue}} \] Mass of tissue: \[ m = 100\, \text{g} = 0.1\, \text{kg} \] ### Step 2: Calculate dose \[ D = \frac{0.101\, \text{J}}{0.1\, \text{kg}} = 1.01\, \text{Gy} \] ### **Final answer:** \[ \boxed{ \textbf{Absorbed dose} \approx 1.01\, \text{Gy} } \] --- # **Summary of Results:** | Part | Quantity | Final Answer | |---------|-------------------------|------------------------------| | (a) | Number of electrons emitted | \(\boxed{9.0 \times 10^{11}}\) electrons | | (b) | Total energy deposited | \(\boxed{0.101\, \text{J}}\) | | (c) | Absorbed dose | \(\boxed{1.01\, \text{Gy}}\) | --- **Note:** All calculations are approximate, assuming uniform energy distribution per electron and complete absorption of emitted electrons within the tissue.