A rare isotope of a nuclear material is very​ unstable, decaying at a rate of 16​% each second. Find how much isotope remains 11 seconds after 8 grams of the isotope is created. Use y equals 8 left parenthesis 2.7 right parenthesis Superscript negative 0.16 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 provided:

\[ y = 8 \cdot (2.7)^{-0.16t} \]

where \( t \) is the time in seconds, and \( y \) represents the amount of the isotope remaining.

Step 1: Plug in the value of \( t \)

In this case, \( t = 11 \) seconds. Plugging this value into the equation gives:

\[ y = 8 \cdot (2.7)^{-0.16 \cdot 11} \]

Step 2: Calculate the exponent

First, we calculate the exponent:

\[ -0.16 \cdot 11 = -1.76 \]

Step 3: Substitute the exponent into the equation

Now we substitute the value into the equation:

\[ y = 8 \cdot (2.7)^{-1.76} \]

Step 4: Calculate \( (2.7)^{-1.76} \)

Using a calculator (or an exponential function calculator), we find:

\[ (2.7)^{-1.76} \approx 0.1770 \quad \text{(this value may vary slightly depending on rounding)} \]

Step 5: Calculate the remaining amount of isotope

Now, substitute this back into the equation:

\[ y \approx 8 \cdot 0.1770 \approx 1.416 \]

Conclusion

Thus, the amount of the rare isotope remaining after 11 seconds is approximately:

\[ \boxed{1.416} \text{ grams} \]