Question

A certain radioactive isotope has a half-life of 25 years, which equates to a decay rate of 0.0278. If there are currently 200 grams of the isotope, determine how much of the isotope will remain after 10 years.(1 point)

Answer 1

The amount of a radioactive isotope remaining after a certain time can be calculated using the formula:

\[
N(t) = N_0 \times e^{-kt}
\]

Where:
- \( N(t) \) is the remaining quantity of the substance after time \( t \).
- \( N_0 \) is the initial quantity of the substance.
- \( k \) is the decay constant.
- \( t \) is the time elapsed.
- \( e \) is the base of the natural logarithm.

From the problem, we have:

- \( N_0 = 200 \) grams
- The decay rate \( k = 0.0278 \)
- \( t = 10 \) years

Now we can substitute the values into the formula:

\[
N(10) = 200 \times e^{-0.0278 \times 10}
\]

Calculating \( -0.0278 \times 10 \):

\[
-0.0278 \times 10 = -0.278
\]

Now we calculate \( e^{-0.278} \):

\[
e^{-0.278} \approx 0.757 (using a calculator or mathematical tables)
\]

Now substituting this back into the equation:

\[
N(10) \approx 200 \times 0.757
\]
\[
N(10) \approx 151.4 \text{ grams}
\]

Therefore, after 10 years, approximately **151.4 grams** of the radioactive isotope will remain.