Evaluate The Following: 1.C 4,4 2. C7,3+C11,9 3. C - Gauthmath

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Answer\(4.516 \text{ gm}\)ExplanationStep 1: Given the half-life of the radioactive isotope is \(5715\) years. Step 2: Use the formula for radioactive decay \(f(x) = A_0e^{-rt}\), where \(A_0\) is the initial amount, \(r\) is the decay rate, and \(t\) is the time in years. Step 3: Since the half-life means the amount is reduced to half, we have \(\frac{A_0}{2} = A_0e^{-r \times 5715}\). Step 4: Simplify the equation to \(\frac{1}{2} = e^{-r \times 5715}\). Step 5: Take the natural logarithm of both sides to solve for \(r\), which gives \(ln(\frac{1}{2}) = -r \times 5715\). Step 6: Calculate \(r\) using the properties of logarithms, \(ln(\frac{1}{2}) = -0.693\), so \(r = \frac{0.693}{5715} = 0.0001213\). Step 7: To find the amount after \(1000\) years, use the decay formula \(A = A_0e^{-rt}\). Step 8: Substitute \(r = 0.0001213\) and \(t = 1000\) into the formula to get \(A = 40e^{-0.0001213 \times 1000}\). Step 9: Calculate \(A\) to find the remaining amount after \(1000\) years, which is \(A = 40e^{-0.1213} = 40 \times 0.88858 = 35.5432 \text{ gm}\). Step 10: Since the initial amount was \(40 \text{ gm}\), the initial quantity is \(40 \text{ gm}\). However, there seems to be a discrepancy in the final step where the amount after \(1000\) years is calculated as \(35.5432 \text{ gm}\), but the initial quantity is given as \(4.516 \text{ gm}\). The correct calculation should be \(40 \times 0.88858 = 35.5432 \text{ gm}\), which is not equal to \(4.516 \text{ gm}\). The correct initial quantity should be \(40 \text{ gm}\), not \(4.516 \text{ gm}\).Click to rate:45(60 votes)Search questionBy textBy image/screenshotDrop your file here orClick Hereto upload