   Chapter 4.6, Problem 15E ### 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 0.3 gram 15. 246 Cm 4730 2 grams

To determine

To calculate: The initial amount of C246m and the amount left after 1000 years, if the half-life of C246m is 4730 years and the amount after 10,000 years is 2 grams.

Explanation

Given Information:

The provided half-life of C246m is 4730 years and the amount after 10,000 years is 2 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 C246m is 4730 years and the amount after 10,000 years is 2 grams.

Now, at t=10,000, the amount of C246m is,

y=2

Substitute t=10,000 and y=2 in the equation y=Cekt.

2=Cek×100002=Ce10000k

The half-life of C246m is 4730 years, that is, at t=4730 the initial amount reduces to half so,

y=C2

Substitute t=4730 and y=C2 in the equation y=Cekt.

C2=Cek×4730C2C=e4730k12=e4730k

Take natural log on both the sides and use the property of natural logarithm,

ln(e4730k)=ln(12)4730k(lne)=ln(12)4730k=ln(12)k=ln(12)4730

Hence, the value of k is ln(12)4730

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