QuestionApril 29, 2025

16 g of a radioactive isotope are present at 2 :00 PM and 1 g remain at 2:00 PM the following day.If the radioactive isotope decay as an exponential model , what's the half-life in hours?

16 g of a radioactive isotope are present at 2 :00 PM and 1 g remain at 2:00 PM the following day.If the radioactive isotope decay as an exponential model , what's the half-life in hours?
16 g of a radioactive isotope are present at 2 :00 PM and 1 g remain at 2:00 PM the following day.If the radioactive isotope decay as an exponential model , what's the half-life in hours?

Solution
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Answer

The half-life of the radioactive isotope is 6 hours. Explanation 1. Determine the decay constant The exponential decay model is given by \( N(t) = N_0 e^{-\lambda t} \), where \( N_0 \) is the initial quantity, ( N(t) ) is the remaining quantity after time ( t ), and ( \lambda ) is the decay constant. Here, \( N_0 = 16 \) g, ( N(t) = 1 ) g, and ( t = 24 ) hours (from 2:00 PM one day to 2:00 PM the next day). We can solve for ( \lambda ) using the equation: \[ 1 = 16 e^{-\lambda \times 24} \] Taking the natural logarithm of both sides gives: \[ \ln(1) = \ln(16) - \lambda \times 24 \] Since \(\ln(1) = 0\), we have: \[ 0 = \ln(16) - \lambda \times 24 \] Solving for ( ( \(\lambda\) ) ): \[ \lambda = \frac{\ln(16)}{24} \] 2. Calculate the half-life The half-life \( T_{1/2} \) is related to the decay constant by the formula \( T_{1/2} = \frac{\ln(2)}{\lambda} \). Substituting the expression for ( ( \(\lambda\) ) ) from Step 1: \[ T_{1/2} = \frac{\ln(2)}{\frac{\ln(16)}{24}} \] Simplifying this expression: \[ T_{1/2} = \frac{24 \ln(2)}{\ln(16)} \] Since \(\ln(16) = 4 \ln(2)\), we substitute: \[ T_{1/2} = \frac{24 \ln(2)}{4 \ln(2)} = \frac{24}{4} = 6 \]

Explanation

1. Determine the decay constant<br /> The exponential decay model is given by \( N(t) = N_0 e^{-\lambda t} \), where \( N_0 \) is the initial quantity, ( N(t) ) is the remaining quantity after time ( t ), and ( \lambda ) is the decay constant. Here, \( N_0 = 16 \) g, ( N(t) = 1 ) g, and ( t = 24 ) hours (from 2:00 PM one day to 2:00 PM the next day). We can solve for ( \lambda ) using the equation:<br /><br />\[<br />1 = 16 e^{-\lambda \times 24}<br />\]<br /><br />Taking the natural logarithm of both sides gives:<br /><br />\[<br />\ln(1) = \ln(16) - \lambda \times 24<br />\]<br /><br />Since \(\ln(1) = 0\), we have:<br /><br />\[<br />0 = \ln(16) - \lambda \times 24<br />\]<br /><br />Solving for ( \(\lambda\) ):<br /><br />\[<br />\lambda = \frac{\ln(16)}{24}<br />\]<br /><br />2. Calculate the half-life<br /> The half-life \( T_{1/2} \) is related to the decay constant by the formula \( T_{1/2} = \frac{\ln(2)}{\lambda} \). Substituting the expression for ( \(\lambda\) ) from Step 1:<br /><br />\[<br />T_{1/2} = \frac{\ln(2)}{\frac{\ln(16)}{24}}<br />\]<br /><br />Simplifying this expression:<br /><br />\[<br />T_{1/2} = \frac{24 \ln(2)}{\ln(16)}<br />\]<br /><br />Since \(\ln(16) = 4 \ln(2)\), we substitute:<br /><br />\[<br />T_{1/2} = \frac{24 \ln(2)}{4 \ln(2)} = \frac{24}{4} = 6<br />\]
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