   Chapter 4.6, Problem 16E ### 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 16. 229 Th 7340 0.3 gram

To determine

To calculate: The initial amount of T229h and the amount left after 1000 years, if the half-life of T229h is 7340 years and the amount after 10,000 years is 0.3 grams.

Explanation

Given Information:

The provided half-life of T229h is 7340 years and the amount after 10,000 years is 0.3 grams.

Formula used:

Exponential growth and decay:

If the rate of change of a positive quantity y with respect to time is proportional to the amount of quantity present at any time t, that is dydt=ky, then y is given by the equation, y=Cekt, where C is the value of the quantity at time t=0 and k is the constant of proportionality.

If k>0 then there is exponential growth and when k<0 then there is exponential decay.

Calculation:

Consider the provided information that the half-life of T229h is 7340 years and the amount after 10,000 years is 0.3 grams.

Observe that here at t=10,000, the amount of T229h is y=0.3.

Substitute t=10,000 and y=0.3 in the equation y=Cekt.

0.3=Cek×100000.3=Ce10000k

Also, the half-life of T229h is 7340 years, that is, at t=7340 the initial amount reduces to half so,

y=C2

Substitute t=7340 and y=C2 in the equation y=Cekt.

C2=Cek×7340C2C=e7340k12=e7340k

Take natural log on both the sides,

ln(e7340k)=ln(12)7340k(lne)=ln(12)7340k=ln(12)k=ln(12)7340

Hence, the value of k is ln(12)7340

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