   Chapter 4.6, Problem 11E ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

#### Solutions

Chapter
Section ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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# Modeling Radioactive Decay In Exercises 11-16, complete the table for each radioactive isotope. See Example 1. Isotope Half-life (in years) Initialquantity Amountafter1000years Amountafter10,000years 11. 226 Ra 1599 10 grams

To determine

To calculate: The amount of R226a left after 1000 years and 10,000 years, if the half-life of R226a is 1599 years and the initial quantity is 10 grams.

Explanation

Given Information:

The provided half-life of R226a is 1599 years and the initial quantity is 10 grams.

Formula used:

The equation for a positive quantity y whose rate of change with respect to time is proportional to the amount of quantity is,

y=Cekt

Where, C is the amount of quantity, t is the any time and k is the proportionality constant.

The exponential growth when k>0 and the exponential growth when k<0.

The property of natural logarithm,

lnxn=nlnx

Calculation:

Consider the provided information that the half-life of R226a is 1599 years and the initial quantity is 10 grams.

Now, at t=0, the initial amount y=10.

Therefore, the value of C is 10.

Substitute C=10 in the equation y=Cekt.

y=10ekt

The half-life of R226a is 1599 years, that is, at t=1599 the initial amount reduces to half so,

y=102=5

Substitute t=1599 and y=5 in the equation y=10ekt.

5=10ek(1599)e1599k=510

Take natural log on both the sides and use the property of logarithm, lnxn=nlnx.

ln(e1599k)=ln(510)1599k(lne)=ln(12)1599k=ln(12)k=ln(12)1599

Therefore, the value of k is ln(12)1599

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