Question

1. americium - 241 is a synthetic isotope used in ionization - type smoke detectors. this radioactive source decays over time. the amount, a micrograms, of americium - 241 present in a detector t years after being produced can be modelled by ( a = 0.3e^{-kt} ) where ( tgeq0 ) and k is a positive constant. (a) determine the amount of americium - 241 present in a new smoke detector immediately after it is produced. it is found that there are 0.295223 micrograms of americium - 241 remaining in a detector after 10 years. (b) find the value of k to four significant figures. (c) find the number of whole years it takes a new detector to release half of the original amount of americium - 241.

Explanation:

Step1: Find the amount when \( t = 0 \)

Substitute \( t = 0 \) into the formula \( A=0.3e^{-kt} \).
Since \( e^{0}=1 \), we have \( A = 0.3e^{0}=0.3\times1 = 0.3 \) micrograms.

Step2: Solve for \( k \) when \( t = 10 \) and \( A=0.295223 \)

Substitute into \( A = 0.3e^{-kt} \), so \( 0.295223=0.3e^{-10k} \).
First, divide both sides by \( 0.3 \): \( \frac{0.295223}{0.3}=e^{-10k} \), which gives \( 0.984077 = e^{-10k} \).
Take the natural logarithm of both sides: \( \ln(0.984077)=-10k \).
Then \( k=-\frac{\ln(0.984077)}{10} \).
Using a calculator, \( \ln(0.984077)\approx - 0.016 \), so \( k=\frac{0.016}{10}=0.0016 \).

Step3: Find \( t \) when \( A = 0.15 \) (half of \( 0.3 \))

Substitute into \( A = 0.3e^{-kt} \), so \( 0.15=0.3e^{-0.0016t} \).
Divide both sides by \( 0.3 \): \( 0.5=e^{-0.0016t} \).
Take the natural logarithm: \( \ln(0.5)=-0.0016t \).
Then \( t=-\frac{\ln(0.5)}{0.0016} \).
Using a calculator, \( \ln(0.5)\approx - 0.6931 \), so \( t=\frac{0.6931}{0.0016}\approx433.19 \).
The number of whole years is \( 433 \).

Answer:

(a) \( 0.3 \) micrograms.
(b) \( k = 0.0016 \).
(c) \( 433 \) years.