QuestionJune 24, 2025

If the half-life of vert -131 is 4.1 days, how many days would it take for only 3.2times 10^-2 grams of vert -131 to be left in Mr. Miller's body if Dr. Goodie started him on an initial dose of 6.5 grams? (Round your answer to two decimal points) square

If the half-life of vert -131 is 4.1 days, how many days would it take for only 3.2times 10^-2 grams of vert -131 to be left in Mr. Miller's body if Dr. Goodie started him on an initial dose of 6.5 grams? (Round your answer to two decimal points) square
If the half-life of vert -131 is 4.1 days, how many days would it take for only 3.2times 10^-2 grams of vert -131 to be left in Mr. Miller's body if Dr. Goodie started him on an initial dose of 6.5 grams? (Round your answer to two decimal points) square

Solution
4.3(136 votes)

Answer

28.72 days Explanation 1. Use the half-life formula The formula for decay is N(t) = N_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}, where N(t) is the remaining quantity, N_0 is the initial quantity, T_{1/2} is the half-life, and t is the time. 2. Set up the equation Given N_0 = 6.5 grams, N(t) = 3.2 \times 10^{-2} grams, and T_{1/2} = 4.1 days, solve for t: 3.2 \times 10^{-2} = 6.5 \left(\frac{1}{2}\right)^{\frac{t}{4.1}}. 3. Solve for t Divide both sides by 6.5: \frac{3.2 \times 10^{-2}}{6.5} = \left(\frac{1}{2}\right)^{\frac{t}{4.1}}. Take the logarithm of both sides: \log\left(\frac{3.2 \times 10^{-2}}{6.5}\right) = \frac{t}{4.1} \log\left(\frac{1}{2}\right). Solve for t: t = 4.1 \cdot \frac{\log\left(\frac{3.2 \times 10^{-2}}{6.5}\right)}{\log\left(\frac{1}{2}\right)}.

Explanation

1. Use the half-life formula<br /> The formula for decay is $N(t) = N_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}$, where $N(t)$ is the remaining quantity, $N_0$ is the initial quantity, $T_{1/2}$ is the half-life, and $t$ is the time.<br />2. Set up the equation<br /> Given $N_0 = 6.5$ grams, $N(t) = 3.2 \times 10^{-2}$ grams, and $T_{1/2} = 4.1$ days, solve for $t$: $3.2 \times 10^{-2} = 6.5 \left(\frac{1}{2}\right)^{\frac{t}{4.1}}$.<br />3. Solve for t<br /> Divide both sides by 6.5: $\frac{3.2 \times 10^{-2}}{6.5} = \left(\frac{1}{2}\right)^{\frac{t}{4.1}}$. Take the logarithm of both sides: $\log\left(\frac{3.2 \times 10^{-2}}{6.5}\right) = \frac{t}{4.1} \log\left(\frac{1}{2}\right)$. Solve for $t$: $t = 4.1 \cdot \frac{\log\left(\frac{3.2 \times 10^{-2}}{6.5}\right)}{\log\left(\frac{1}{2}\right)}$.
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