question 21 of 30 how much of the original amount of an isotope is present after a period of four half…

question 21 of 30 how much of the original amount of an isotope is present after a period of four half - lives? a. sixteen times the original amount b. eight times the original amount c. one - sixteenth of the original amount d. one - eighth of the original amount

question 21 of 30 how much of the original amount of an isotope is present after a period of four half - lives? a. sixteen times the original amount b. eight times the original amount c. one - sixteenth of the original amount d. one - eighth of the original amount

Answer

Answer:

C. One - sixteenth of the original amount

Explanation:

Step1: Recall half - life formula

The amount of a substance remaining after $n$ half - lives is given by $A = A_0\times(\frac{1}{2})^n$, where $A_0$ is the initial amount and $n$ is the number of half - lives.

Step2: Substitute $n = 4$

When $n = 4$, we have $A=A_0\times(\frac{1}{2})^4$.

Step3: Calculate $(\frac{1}{2})^4$

$(\frac{1}{2})^4=\frac{1}{2\times2\times2\times2}=\frac{1}{16}$. So $A=\frac{1}{16}A_0$, which means one - sixteenth of the original amount remains.