QuestionJune 8, 2025

A radioactive substance decays exponentially. A scientist begins with 140 milligrams of a radioactive substance. After 24 hours 70 mg of the substance remains. How many milligrams will remain after 27 hours? square mg Give your answer accurate to at least one decimal place Question Help: Video

A radioactive substance decays exponentially. A scientist begins with 140 milligrams of a radioactive substance. After 24 hours 70 mg of the substance remains. How many milligrams will remain after 27 hours? square mg Give your answer accurate to at least one decimal place Question Help: Video
A radioactive substance decays exponentially. A scientist begins with 140 milligrams of a radioactive substance. After 24 hours 70 mg of the substance remains. How many milligrams will remain after 27 hours? square mg Give your answer accurate to at least one decimal place Question Help: Video

Solution
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Answer

61.9 mg Explanation 1. Determine the decay constant Use the formula for exponential decay: N(t) = N_0 e^{-\lambda t}. Given N_0 = 140 mg, N(24) = 70 mg, and t = 24 hours. Solve for \lambda: \[ 70 = 140 e^{-24\lambda} \] \[ \frac{1}{2} = e^{-24\lambda} \] Take the natural logarithm: \[ -24\lambda = \ln\left(\frac{1}{2}\right) \] \[ \lambda = -\frac{\ln(0.5)}{24} \] 2. Calculate remaining substance after 27 hours Use the decay constant \lambda to find N(27): \[ N(27) = 140 e^{-\lambda \cdot 27} \] Substitute \lambda from Step 1: \[ N(27) = 140 e^{\frac{\ln(0.5)}{24} \cdot 27} \]

Explanation

1. Determine the decay constant<br /> Use the formula for exponential decay: $N(t) = N_0 e^{-\lambda t}$. Given $N_0 = 140$ mg, $N(24) = 70$ mg, and $t = 24$ hours. Solve for $\lambda$: <br />\[ 70 = 140 e^{-24\lambda} \]<br />\[ \frac{1}{2} = e^{-24\lambda} \]<br />Take the natural logarithm:<br />\[ -24\lambda = \ln\left(\frac{1}{2}\right) \]<br />\[ \lambda = -\frac{\ln(0.5)}{24} \]<br /><br />2. Calculate remaining substance after 27 hours<br /> Use the decay constant $\lambda$ to find $N(27)$:<br />\[ N(27) = 140 e^{-\lambda \cdot 27} \]<br />Substitute $\lambda$ from Step 1:<br />\[ N(27) = 140 e^{\frac{\ln(0.5)}{24} \cdot 27} \]
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