Solved on Feb 09, 2024

Erbium isotope with 12-hour half-life, initial 22g. Find: a) A(t)A(t) function, b) decay rate A(t)A'(t), c) decay rate at 11 hours.
a) A(t)=22e0.0577tA(t) = 22e^{-0.0577t} grams b) A(t)=1.2694e0.0577tA'(t) = -1.2694e^{-0.0577t} grams/hour c) A(11)=0.2069A'(11) = 0.2069 grams/hour

STEP 1

Assumptions
1. The initial amount of the erbium isotope is 22 grams.
2. The half-life of the erbium isotope is 12 hours.
3. The decay of the substance can be modeled by an exponential function.
4. The rate of decay is the derivative of the amount of substance remaining with respect to time.

STEP 2

To write the exponential function that relates the amount of substance remaining, A(t)A(t), as a function of time tt, we use the formula for exponential decay:
A(t)=A0ektA(t) = A_0 \cdot e^{kt}
where A0A_0 is the initial amount of the substance, kk is the decay constant, and tt is the time in hours.

STEP 3

Since we know the half-life of the substance is 12 hours, we can find the decay constant kk using the half-life formula:
12A0=A0ekhalf-life\frac{1}{2}A_0 = A_0 \cdot e^{k \cdot \text{half-life}}

STEP 4

Simplify the half-life formula by dividing both sides by A0A_0:
12=ek12\frac{1}{2} = e^{k \cdot 12}

STEP 5

Take the natural logarithm of both sides to solve for kk:
ln(12)=k12\ln\left(\frac{1}{2}\right) = k \cdot 12

STEP 6

Calculate the value of kk:
k=ln(12)12k = \frac{\ln\left(\frac{1}{2}\right)}{12}

STEP 7

Substitute the value of kk into the exponential decay formula:
A(t)=22e(ln(12)12)tA(t) = 22 \cdot e^{\left(\frac{\ln\left(\frac{1}{2}\right)}{12}\right)t}

STEP 8

Simplify the function by combining the constants:
A(t)=22e(ln(2)12)tA(t) = 22 \cdot e^{\left(-\frac{\ln(2)}{12}\right)t}
This is the exponential function that relates the amount of substance remaining, A(t)A(t), as a function of time tt.

STEP 9

To find the rate at which the substance is decaying after tt hours, we need to find the derivative of A(t)A(t) with respect to tt. This is denoted as A(t)A'(t).

STEP 10

Differentiate the function A(t)A(t) with respect to tt:
A(t)=ddt(22e(ln(2)12)t)A'(t) = \frac{d}{dt}\left(22 \cdot e^{\left(-\frac{\ln(2)}{12}\right)t}\right)

STEP 11

Apply the chain rule to differentiate the exponential function:
A(t)=22(ln(2)12)e(ln(2)12)tA'(t) = 22 \cdot \left(-\frac{\ln(2)}{12}\right) \cdot e^{\left(-\frac{\ln(2)}{12}\right)t}

STEP 12

Simplify the expression for the rate of decay:
A(t)=22ln(2)12e(ln(2)12)tA'(t) = -\frac{22 \cdot \ln(2)}{12} \cdot e^{\left(-\frac{\ln(2)}{12}\right)t}
This is the rate at which the substance is decaying after tt hours.

STEP 13

To find the rate of decay at 11 hours, substitute t=11t = 11 into the rate of decay function A(t)A'(t):
A(11)=22ln(2)12e(ln(2)12)11A'(11) = -\frac{22 \cdot \ln(2)}{12} \cdot e^{\left(-\frac{\ln(2)}{12}\right) \cdot 11}

STEP 14

Calculate the value of A(11)A'(11):
A(11)=22ln(2)12e(ln(2)12)11A'(11) = -\frac{22 \cdot \ln(2)}{12} \cdot e^{\left(-\frac{\ln(2)}{12}\right) \cdot 11}

STEP 15

Use a calculator to compute the value of A(11)A'(11) and round to four decimal places:
A(11)22ln(2)12e(ln(2)12)111.6395A'(11) \approx -\frac{22 \cdot \ln(2)}{12} \cdot e^{\left(-\frac{\ln(2)}{12}\right) \cdot 11} \approx -1.6395
grams per hour (rounded to four decimal places).
The exponential function that relates the amount of substance remaining is: A(t)=22e(ln(2)12)t A(t) = 22 \cdot e^{\left(-\frac{\ln(2)}{12}\right)t} grams.
The rate at which the substance is decaying after tt hours is: A(t)=22ln(2)12e(ln(2)12)t A^{\prime}(t) = -\frac{22 \cdot \ln(2)}{12} \cdot e^{\left(-\frac{\ln(2)}{12}\right)t} grams per hour.
The rate of decay at 11 hours is approximately: A(11)1.6395 A^{\prime}(11) \approx -1.6395 grams per hour.

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