21) The beta decay of cesium- 137 has a half-life of 30.0 years. How many years must pass to reduce a 30 mg sample of cesium 137 to 5.2 mg ? A. 38 years B. 0.040 years C. 25 years D. 76 years

To solve this problem, we can use the concept of half-life and the formula for radioactive decay. Given: - Half-life of cesium-137 = 30 years - Initial mass of cesium-137 = 30 mg - Final mass of cesium-137 = 5.2 mg We can use the formula for radioactive decay: \[ A = A_0 \times \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}} \] Where: - \( A \) is the final mass of cesium-137 - \( A_0 \) is the initial mass of cesium-137 - \( t \) is the time that has passed - \( t_{1/2} \) is the half-life of cesium-137 Substitute the given values into the formula: \[ 5.2 = 30 \times \left( \frac{1}{2} \right)^{\frac{t}{30}} \] Now, we can solve for \( t \) to find out how many years must pass to reduce the sample of cesium-137 to 5.2 mg. Solve the equation by following steps: - step0: Solve for \(t\): \(5.2=30\left(\frac{1}{2}\right)^{\frac{t}{30}}\) - step1: Swap the sides: \(30\left(\frac{1}{2}\right)^{\frac{t}{30}}=5.2\) - step2: Convert the expressions: \(30\left(\frac{1}{2}\right)^{\frac{t}{30}}=\frac{26}{5}\) - step3: Multiply by the reciprocal: \(30\left(\frac{1}{2}\right)^{\frac{t}{30}}\times \frac{1}{30}=\frac{26}{5}\times \frac{1}{30}\) - step4: Multiply: \(\left(\frac{1}{2}\right)^{\frac{t}{30}}=\frac{13}{75}\) - step5: Take the logarithm of both sides: \(\log_{\frac{1}{2}}{\left(\left(\frac{1}{2}\right)^{\frac{t}{30}}\right)}=\log_{\frac{1}{2}}{\left(\frac{13}{75}\right)}\) - step6: Evaluate the logarithm: \(\frac{t}{30}=\log_{\frac{1}{2}}{\left(\frac{13}{75}\right)}\) - step7: Cross multiply: \(t=30\log_{\frac{1}{2}}{\left(\frac{13}{75}\right)}\) - step8: Simplify: \(t=30\log_{2}{\left(75\right)}-30\log_{2}{\left(13\right)}\) The time that must pass to reduce a 30 mg sample of cesium-137 to 5.2 mg is given by the equation: \[ t = 30 \log_{2}(75) - 30 \log_{2}(13) \] Now, we can calculate the value of \( t \) to determine the answer. Calculate the value by following steps: - step0: Calculate: \(30\log_{2}{\left(75\right)}-30\log_{2}{\left(13\right)}\) - step1: Rewrite the expression: \(\log_{2}{\left(75^{30}\right)}-\log_{2}{\left(13^{30}\right)}\) - step2: Use the logarithm product rule: \(\log_{2}{\left(\frac{75^{30}}{13^{30}}\right)}\) - step3: Write in exponential form: \(\log_{2}{\left(\left(\frac{75}{13}\right)^{30}\right)}\) - step4: Calculate: \(30\log_{2}{\left(\frac{75}{13}\right)}\) The time that must pass to reduce a 30 mg sample of cesium-137 to 5.2 mg is approximately 75.85 years. Therefore, the correct answer is: D. 76 years