Question
Consider the following scenario: A sample of radioactive material has a decay constant of 0.01 per day. If there are initially 100 grams of the material, what equation would you use to determine how much will remain after 50 days?
1)y=50(0.99)^100
2)y=50(0.01)^100
3)y=100(0.99)^50
4)y=100(0.01)^50
1)y=50(0.99)^100
2)y=50(0.01)^100
3)y=100(0.99)^50
4)y=100(0.01)^50
Answers
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Answered by
GPT-4o mini
To determine the remaining amount of a radioactive substance after a given period, you can use the exponential decay formula:
\[ N(t) = N_0 e^{-\lambda t} \]
Where:
- \( N(t) \) is the amount remaining after time \( t \),
- \( N_0 \) is the initial amount of the substance,
- \( \lambda \) is the decay constant,
- \( t \) is the time in the same units as the decay constant.
In your scenario:
- The initial amount \( N_0 = 100 \) grams,
- The decay constant \( \lambda = 0.01 \) per day,
- The time \( t = 50 \) days.
Alternatively, a simplified version of the equation can be expressed using the half-life or the property that the remaining amount can also be calculated as:
\[ N(t) = N_0 (1 - p)^t \]
where \( p = e^{-\lambda} \) is the proportion remaining after one time period (in this case, one day). In your case, since \( 1 - p \) is very close to \( 0.01 \) per day:
Using this simplified form, we want to calculate:
\[ y = 100(0.99)^{50} \]
Thus, the correct equation from the options provided, where you replace the base with \( 0.99 \) raised to the power of the number of days, looks like:
**Answer: 3) \( y = 100(0.99)^{50} \)**
\[ N(t) = N_0 e^{-\lambda t} \]
Where:
- \( N(t) \) is the amount remaining after time \( t \),
- \( N_0 \) is the initial amount of the substance,
- \( \lambda \) is the decay constant,
- \( t \) is the time in the same units as the decay constant.
In your scenario:
- The initial amount \( N_0 = 100 \) grams,
- The decay constant \( \lambda = 0.01 \) per day,
- The time \( t = 50 \) days.
Alternatively, a simplified version of the equation can be expressed using the half-life or the property that the remaining amount can also be calculated as:
\[ N(t) = N_0 (1 - p)^t \]
where \( p = e^{-\lambda} \) is the proportion remaining after one time period (in this case, one day). In your case, since \( 1 - p \) is very close to \( 0.01 \) per day:
Using this simplified form, we want to calculate:
\[ y = 100(0.99)^{50} \]
Thus, the correct equation from the options provided, where you replace the base with \( 0.99 \) raised to the power of the number of days, looks like:
**Answer: 3) \( y = 100(0.99)^{50} \)**
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