A rare isotope of a nuclear material is very​ unstable, decaying at a rate of 15​% each second. Find how much isotope remains 11 seconds after 4 grams of the isotope is created. Use y equals 4 left parenthesis 2.7 right parenthesis Superscript negative 0.15 t​, where t represents time in seconds.

Answer 1

To find out how much of the rare isotope remains after 11 seconds, we can use the formula given:

\[ y = 4 \times (2.7)^{-0.15t} \]

Here, \( y \) represents the remaining amount of the isotope, \( 4 \) grams is the initial amount, \( t \) is the time in seconds, and the decay rate is represented by the exponent \(-0.15t\).

Substituting \( t = 11 \) seconds into the equation:

\[ y = 4 \times (2.7)^{-0.15 \times 11} \]

First, we calculate the exponent:

\[ -0.15 \times 11 = -1.65 \]

Now substitute this back into the equation:

\[ y = 4 \times (2.7)^{-1.65} \]

Next, we need to calculate \( (2.7)^{-1.65} \).

Calculating \( (2.7)^{-1.65} \):

\[ (2.7)^{-1.65} \approx 0.1842 \quad (\text{using a calculator}) \]

Now substitute this value back into the equation for \( y \):

\[ y = 4 \times 0.1842 \approx 0.7368 \]

Thus, after 11 seconds, approximately 0.7368 grams of the isotope remains. Rounded to two decimal places, that would be:

\[ \text{Remaining isotope} \approx 0.74 \text{ grams} \]