Question

question 5 of 10
the function ( q(t)=q_{0} e^{-k t} ) may be used to model radioactive decay. ( q ) represents the quantity remaining after ( t ) years; ( k ) is the decay constant. what is the decay constant for plutonium - 240 if its half - life is 6,300 years?
a. ( ln (0.5) / ln (6,300) )
b. ( -ln (0.5) / 6,300 )
c. ( 240 ln (0.5) / 6,300 )
d. ( ln (0.5) / 240 )

Explanation:

Step1: Set up half-life condition

At half-life \( t = 6300 \), \( Q = 0.5Q_0 \). Substitute into \( Q(t) = Q_0 e^{-kt} \):
\( 0.5Q_0 = Q_0 e^{-k(6300)} \)

Step2: Simplify equation

Divide both sides by \( Q_0 \):
\( 0.5 = e^{-6300k} \)

Step3: Take natural log of both sides

\( \ln(0.5) = -6300k \)

Step4: Solve for k

Rearrange to isolate k:
\( k = -\frac{\ln(0.5)}{6300} \)

Answer:

B. -[ln(0.5)/6,300]