   Chapter 4.6, Problem 14E ### 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 14. 63 Ni 96 1.5 grams

To determine

To calculate: The initial amount of N63i and the amount left after 10,000 years, if the half-life of N63i is 96 years and the amount after 1000 years is 1.5 grams.

Explanation

Given Information:

The provided half-life of N63i is 96 years and the amount after 1000 years is 1.5 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 N63i is 96 years and the amount after 1000 years is 1.5 grams.

Now, at t=1000, the amount of N63i is y=1.5.

Substitute t=1000 and y=1.5 in the equation y=Cekt.

1.5=Cek×10001.5=Ce1000k

The half-life of N63i is 96 years, that is, at t=96 the initial amount reduces to half so,

y=C2

Substitute t=96 and y=C2 in the equation y=Cekt.

C2=Cek×96C2C=e96k12=e96k

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

ln(e96k)=ln(12)96k(lne)=ln(12)96k=ln(12)k=ln(12)96

Therefore, the value of k is ln(12)96

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