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A radioactive sample decays to 10% of its initial amount in 4600 minutes. The rate constant of this process is __________ hour-1. (Round off to two decimal places)
Correct answer is between '0.02,0.04'. Can you explain this answer?
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A radioactive sample decays to 10% of its initial amount in 4600 minut...
Given:
- Initial amount of radioactive sample
- Final amount of radioactive sample = 10% of the initial amount
- Time taken for decay = 4600 minutes
We need to find the rate constant of this process in hour-1.
Formula:
The rate of radioactive decay is given by the first-order kinetics equation:
ln(N/N0) = -kt
where
- N is the final amount of radioactive sample
- N0 is the initial amount of radioactive sample
- k is the rate constant of the process
- t is the time taken for decay
Calculations:
1. Convert time from minutes to hours:
4600 minutes = 4600/60 hours = 76.67 hours
2. Given that the final amount of radioactive sample is 10% of the initial amount, we can write:
N/N0 = 0.1
Taking natural logarithm on both sides of the equation, we get:
ln(N/N0) = ln(0.1)
3. Substituting the given values in the first-order kinetics equation, we get:
ln(0.1) = -k x 76.67
4. Solving for k:
k = -ln(0.1)/76.67
k = 0.03133 hour-1 (rounded off to two decimal places)
Answer:
The rate constant of this radioactive decay process is 0.03 hour-1, which is between 0.02 and 0.04 hour-1 as given in the question.
Note: The rate constant is a measure of the speed of the radioactive decay process. A higher value of k means a faster decay process.
- Initial amount of radioactive sample
- Final amount of radioactive sample = 10% of the initial amount
- Time taken for decay = 4600 minutes
We need to find the rate constant of this process in hour-1.
Formula:
The rate of radioactive decay is given by the first-order kinetics equation:
ln(N/N0) = -kt
where
- N is the final amount of radioactive sample
- N0 is the initial amount of radioactive sample
- k is the rate constant of the process
- t is the time taken for decay
Calculations:
1. Convert time from minutes to hours:
4600 minutes = 4600/60 hours = 76.67 hours
2. Given that the final amount of radioactive sample is 10% of the initial amount, we can write:
N/N0 = 0.1
Taking natural logarithm on both sides of the equation, we get:
ln(N/N0) = ln(0.1)
3. Substituting the given values in the first-order kinetics equation, we get:
ln(0.1) = -k x 76.67
4. Solving for k:
k = -ln(0.1)/76.67
k = 0.03133 hour-1 (rounded off to two decimal places)
Answer:
The rate constant of this radioactive decay process is 0.03 hour-1, which is between 0.02 and 0.04 hour-1 as given in the question.
Note: The rate constant is a measure of the speed of the radioactive decay process. A higher value of k means a faster decay process.
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