Question

Cobalt-60 has a half-life of approximately five years. Which of the following equations correctly models the amount of cobalt-60 remaining after t years?(1 point)
Responses

P(t)=P0e−0.1386t
upper P left parenthesis t right parenthesis equals upper P subscript 0 baseline e superscript negative 0.1386 t baseline

P(t)=P0e−0.6931t
upper P left parenthesis t right parenthesis equals upper P subscript 0 baseline e superscript negative 0.6931 t baseline

P(t)=P0e0.1386t
upper P left parenthesis t right parenthesis equals upper P subscript 0 baseline e superscript 0.1386 t baseline

P(t)=P0e0.6931t

Answer 1

To model the decay of a substance like cobalt-60, which has a half-life of 5 years, we can use the exponential decay formula:

\[ P(t) = P_0 e^{-kt} \]

where:
- \( P(t) \) is the amount of substance remaining after time \( t \),
- \( P_0 \) is the initial amount of the substance,
- \( k \) is the decay constant,
- \( t \) is the time in years.

The decay constant \( k \) can be found from the half-life \( t_{1/2} \) using the formula:

\[ k = \frac{\ln(2)}{t_{1/2}} \]

For cobalt-60 with a half-life of 5 years:

\[ k = \frac{\ln(2)}{5} \]

Calculating \( \ln(2) \):

\[ \ln(2) \approx 0.6931 \]

So:

\[ k \approx \frac{0.6931}{5} \approx 0.1386 \]

Thus, the equation that models the amount of cobalt-60 remaining after \( t \) years is:

\[ P(t) = P_0 e^{-0.1386t} \]

Therefore, the correct option is:

**P(t) = P0e^(-0.1386t)**