   Chapter 4.6, Problem 13E ### 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 13. 239 Pu 24,100 2.1 grams

To determine

To calculate: The initial amount of P239u and the amount left after 10,000 years, if the half-life of P239u is 24,100 years and the amount after 1000 years is 2.1 grams.

Explanation

Given Information:

The provided half-life of P239u is 24,100 years and the amount after 1000 years is 2.1 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 P239u is 24,100 years and the amount after 1000 years is 2.1 grams.

Now, at t=1000, the amount of P239u is y=2.1.

Substitute t=1000 and y=2.1 in the equation y=Cekt.

2.1=Cek×10002.1=Ce1000k

Also the half-life of P239u is 24,100 years, that is, at t=24,100 the initial amount reduces to half so,

y=C2

Substitute t=24,100 and y=C2 in the equation y=Cekt.

C2=Cek×24,100C2C=e24,100k12=e24,100k

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

ln(e24,100k)=ln(12)24,100k(lne)=ln(12)24,100k=ln(12)k=ln(12)24,100

Therefore, the value of k is ln(12)24,100

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