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)

1 answer

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.