EBK PRECALCULUS W/LIMITS
EBK PRECALCULUS W/LIMITS
4th Edition
ISBN: 9781337516853
Author: Larson
Publisher: CENGAGE CO
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Differential Equations
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Chapter 3.5, Problem 24E
To determine

To find : the initial amount of radioactive isotope.

Expert Solution & Answer
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Answer to Problem 24E

The initial amount of radioactive isotope is   A≈0.4117 grams    _

Explanation of Solution

Given information : The radioactive isotope details given in the table

    IsotopeHalf-life (years)Initial QuantityAmount After 1000 Years
    239Pu24100?0.4 g

Concept Involved:

Solving for a variable means getting the variable alone in one side of the equation by undoing whatever operation is done to it.

Formula Used:

For Radioactive Decay, the amount A after t years, with initial amount A0 and decay rater (in decimal form) is given by the formula: A = A0ert

Logarithmic property: lneX=X

Calculation:

    DescriptionSteps
    To find the decay rate, we need to replace 0.5A0 in place of A& t=24100{Because half-life is 24100years} in the equation A = A0ert0.5A0=A0e24100r
    Use symmetric property of equality which states that if a = b then b = a to rewrite the equationA0e24100r=0.5A0
    Divide by A0 on both sides A0e24100rA0=0.5A0A0
    Simplify fraction on both sides of the equatione24100r=0.5
    Take natural logarithm on both sideslne24100r=ln(0.5)

Calculation (Continued):

    DescriptionSteps
    Apply the logarithmic rule lneX=X in left side of the equation 24100r=ln0.5
    Divide by 24100 on both sides 24100r24100=ln0.524100
    Simplify fraction on both sides of the equationr=−2.88×10−5
    Replace 0.4 for A ,t=1000 & −2.88×10−5 for r in the equation A = A0ert0.4 =A0e−2.88×10−5(1000)
    Use symmetric property of equality which states that if a = b then b = a to rewrite the equationA0e−2.88×10−5(1000)=0.4
    Simplify the exponent in the left side of the equationA0e−0.0288=0.4
    Divide by e−0.0288 on both sidesA0e−0.0288e−0.0288=0.4e−0.0288
    Simplify fraction on both sides of the equationA0=0.4117

Conclusion:

    IsotopeHalf-life (years)Initial QuantityAmount After 1000 Years
    14C57150.4117 g0.4 g
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Chapter 3.5, Problem 23E
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Chapter 3.5, Problem 25E

Chapter 3 Solutions

EBK PRECALCULUS W/LIMITS

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