QuestionFind the half-life of a radioactive substance with a decay rate of $4.3\%$ per day using continuous exponential decay.

Studdy Solution

STEP 1

Assumptions1. The decay of the substance is continuous and exponential. . The decay rate parameter is $4.3\%$ per day.
3. We are asked to find the half-life of the substance, which is the time it takes for half of the substance to decay.

STEP 2

The continuous exponential decay model can be represented by the equation $(t) = N0 e^{-kt}$ where- $(t)$ is the amount of substance remaining after time $t$ , - $0$ is the initial amount of the substance, - $k$ is the decay rate parameter, - $e$ is the base of the natural logarithm (approximately equal to2.71828), - $t$ is the time elapsed.

STEP 3

We need to find the time $t$ when $(t) = \frac{1}{2}0$ . This is the definition of half-life. Substituting these values into the decay model equation, we get $\frac{1}{2}0 = N0 e^{-kt}$

STEP 4

We can simplify this equation by dividing both sides by $0$ , which gives $\frac{1}{2} = e^{-kt}$

STEP 5

To solve for $t$ , we can take the natural logarithm of both sides of the equation. This gives $\ln\left(\frac{1}{2}\right) = \ln\left(e^{-kt}\right)$

STEP 6

Using the property of logarithms that $\ln(e^x) = x$ , we can simplify the right side of the equation $\ln\left(\frac{1}{2}\right) = -kt$

STEP 7

To isolate $t$ , we can divide both sides of the equation by $-k$ . This gives $t = \frac{\ln\left(\frac{1}{2}\right)}{-k}$

STEP 8

Now, plug in the given value for $k$ to calculate the half-life. Remember to convert the percentage to a decimal value.
$t = \frac{\ln\left(\frac{1}{2}\right)}{-0.043}$

STEP 9

Calculate the half-life of the substance.
$t \approx16.09\, days$ The half-life of the substance is approximately16.09 days.

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QuestionFind the half-life of a radioactive substance with a decay rate of $4.3\%$ per day using continuous exponential decay.

Studdy Solution

STEP 1

Assumptions1. The decay of the substance is continuous and exponential. . The decay rate parameter is $4.3\%$ per day.
3. We are asked to find the half-life of the substance, which is the time it takes for half of the substance to decay.

STEP 2

STEP 3

We need to find the time $t$ when $(t) = \frac{1}{2}0$ . This is the definition of half-life. Substituting these values into the decay model equation, we get $\frac{1}{2}0 = N0 e^{-kt}$

STEP 4

We can simplify this equation by dividing both sides by $0$ , which gives $\frac{1}{2} = e^{-kt}$

STEP 5

To solve for $t$ , we can take the natural logarithm of both sides of the equation. This gives $\ln\left(\frac{1}{2}\right) = \ln\left(e^{-kt}\right)$

STEP 6

Using the property of logarithms that $\ln(e^x) = x$ , we can simplify the right side of the equation $\ln\left(\frac{1}{2}\right) = -kt$

STEP 7

To isolate $t$ , we can divide both sides of the equation by $-k$ . This gives $t = \frac{\ln\left(\frac{1}{2}\right)}{-k}$

STEP 8

Now, plug in the given value for $k$ to calculate the half-life. Remember to convert the percentage to a decimal value.
$t = \frac{\ln\left(\frac{1}{2}\right)}{-0.043}$

STEP 9

Calculate the half-life of the substance.
$t \approx16.09\, days$ The half-life of the substance is approximately16.09 days.

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QuestionFind the half-life of a radioactive substance with a decay rate of 4.3%4.3\%4.3% per day using continuous exponential decay.

Studdy Solution

STEP 1

Assumptions1. The decay of the substance is continuous and exponential. . The decay rate parameter is 4.3%4.3\%4.3% per day.3. We are asked to find the half-life of the substance, which is the time it takes for half of the substance to decay.

STEP 2

STEP 3

We need to find the time ttt when (t)=120(t) = \frac{1}{2}0(t)=21​0. This is the definition of half-life. Substituting these values into the decay model equation, we get120=N0e−kt\frac{1}{2}0 = N0 e^{-kt}21​0=N0e−kt

STEP 4

We can simplify this equation by dividing both sides by 000, which gives12=e−kt\frac{1}{2} = e^{-kt}21​=e−kt

STEP 5

To solve for ttt, we can take the natural logarithm of both sides of the equation. This givesln⁡(12)=ln⁡(e−kt)\ln\left(\frac{1}{2}\right) = \ln\left(e^{-kt}\right)ln(21​)=ln(e−kt)

STEP 6

Using the property of logarithms that ln⁡(ex)=x\ln(e^x) = xln(ex)=x, we can simplify the right side of the equationln⁡(12)=−kt\ln\left(\frac{1}{2}\right) = -ktln(21​)=−kt

STEP 7

To isolate ttt, we can divide both sides of the equation by −k-k−k. This givest=ln⁡(12)−kt = \frac{\ln\left(\frac{1}{2}\right)}{-k}t=−kln(21​)​

STEP 8

Now, plug in the given value for kkk to calculate the half-life. Remember to convert the percentage to a decimal value.t=ln⁡(12)−0.043t = \frac{\ln\left(\frac{1}{2}\right)}{-0.043}t=−0.043ln(21​)​

STEP 9

Calculate the half-life of the substance.t≈16.09 dayst \approx16.09\, dayst≈16.09daysThe half-life of the substance is approximately16.09 days.

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QuestionFind the half-life of a radioactive substance with a decay rate of $4.3\%$ per day using continuous exponential decay.

Assumptions1. The decay of the substance is continuous and exponential. . The decay rate parameter is $4.3\%$ per day.
3. We are asked to find the half-life of the substance, which is the time it takes for half of the substance to decay.

We need to find the time $t$ when $(t) = \frac{1}{2}0$ . This is the definition of half-life. Substituting these values into the decay model equation, we get $\frac{1}{2}0 = N0 e^{-kt}$

We can simplify this equation by dividing both sides by $0$ , which gives $\frac{1}{2} = e^{-kt}$

To solve for $t$ , we can take the natural logarithm of both sides of the equation. This gives $\ln\left(\frac{1}{2}\right) = \ln\left(e^{-kt}\right)$

Using the property of logarithms that $\ln(e^x) = x$ , we can simplify the right side of the equation $\ln\left(\frac{1}{2}\right) = -kt$

To isolate $t$ , we can divide both sides of the equation by $-k$ . This gives $t = \frac{\ln\left(\frac{1}{2}\right)}{-k}$

Now, plug in the given value for $k$ to calculate the half-life. Remember to convert the percentage to a decimal value.
$t = \frac{\ln\left(\frac{1}{2}\right)}{-0.043}$

Calculate the half-life of the substance.
$t \approx16.09\, days$ The half-life of the substance is approximately16.09 days.