Question: The half-life of phosphorus-32 is approximately 24.3 days. Step 3 of 3: How much of a 3 gram…

The half-life of phosphorus-32 is approximately 24.3 days.

Step 3 of 3: How much of a 3 gram sample of phosphorus-32 would remain after 10 days? Round to three decimal places.

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grams

Solution

To solve this problem, we’ll use the formula for radioactive decay based on half-life: The formula is given by: \[ N(t) = N_0 \times \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}} \] where: \(N(t)\) = the remaining quantity of the substance after time \(t\), \(N_0\) = the initial quantity of the substance, \(T_{1/2}\) = the half-life of the substance. Given: \(N_0 = 3\) grams, \(T_{1/2} = 24.3\) days, \(t = 10\) days. Plug these values into the formula: \[ N(10) = 3 \times \left(\frac{1}{2}\right)^{\frac{10}{24.3}} \] Calculate the exponent: \[ \frac{10}{24.3} \approx 0.4115 \] Evaluate the power of \(\frac{1}{2}\): \[ \left(\frac{1}{2}\right)^{0.4115} \approx 0.7529 \] Calculate \(N(10)\): \[ N(10) = 3 \times 0.7529 \approx 2.259 \] The remaining amount of phosphorus-32 after 10 days is approximately 2.259 grams.