Jjinlov5idn
Answered

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Using the values from Step 1, predict the pattern for the decay of 100 atoms over the course of eight half-life cycles. Round to the nearest whole number of atoms. Record in the appropriate blanks.

[tex]\[ A = \][/tex]
[tex]\[ B = \][/tex]
[tex]\[ C = \][/tex]
[tex]\[ D = \][/tex]
[tex]\[ E = \][/tex]

\begin{tabular}{|c|c|c|}
\hline
\begin{tabular}{c}
Time-Half- \\
Life Cycles, [tex]$n$[/tex]
\end{tabular} & [tex]$0.5^n$[/tex] & \begin{tabular}{c}
Radioactive Atoms \\
(Predicted)
\end{tabular} \\
\hline
Initial & 1 & 100 \\
\hline
1 & 0.5 & A \\
\hline
2 & 0.25 & 25 \\
\hline
3 & 0.125 & B \\
\hline
4 & 0.0625 & 6 \\
\hline
5 & 0.03125 & C \\
\hline
6 & 0.015625 & D \\
\hline
7 & 0.0078125 & 1 \\
\hline
8 & 0.00390625 & E \\
\hline
\end{tabular}

Sagot :

GinnyAnsweridn
Let's start by filling in the values step-by-step:

1. Initial atoms: We begin with 100 atoms.
2. After the 1st half-life cycle (0.5n):
- Half of the initial amount remains.
- Thus, [tex]\(100 / 2 = 50\)[/tex].
- Therefore, [tex]\(A = 50\)[/tex].

3. After the 2nd half-life cycle (0.25n):
- Half of the remaining atoms from the previous cycle.
- [tex]\(50 / 2 = 25\)[/tex].

4. After the 3rd half-life cycle (0.125n):
- Half of the remaining atoms from the previous cycle.
- [tex]\(25 / 2 = 12.5\)[/tex], rounded to the nearest whole number is 12.
- Therefore, [tex]\(B = 12\)[/tex].

5. After the 4th half-life cycle (0.0625n):
- Half of the remaining atoms from the previous cycle.
- [tex]\(12 / 2 = 6\)[/tex].

6. After the 5th half-life cycle (0.03125n):
- Half of the remaining atoms from the previous cycle.
- [tex]\(6 / 2 = 3\)[/tex].
- Therefore, [tex]\(C = 3\)[/tex].

7. After the 6th half-life cycle (0.015625n):
- Half of the remaining atoms from the previous cycle.
- [tex]\(3 / 2 = 1.5\)[/tex], rounded to the nearest whole number is 2.
- Therefore, [tex]\(D = 2\)[/tex].

8. After the 7th half-life cycle (0.0078125n):
- Half of the remaining atoms from the previous cycle.
- [tex]\(2 / 2 = 1\)[/tex].

9. After the 8th half-life cycle (0.00390625n):
- Half of the remaining atoms from the previous cycle.
- [tex]\(1 / 2 = 0.5\)[/tex], rounded to the nearest whole number is 0.
- Therefore, [tex]\(E = 0\)[/tex].

So, putting these values back into the table we get:

[tex]\[ A = 50 \][/tex]
[tex]\[ B = 12 \][/tex]
[tex]\[ C = 3 \][/tex]
[tex]\[ D = 2 \][/tex]
[tex]\[ E = 0 \][/tex]

Here’s the final table:

[tex]\[ \begin{tabular}{|c|c|c|} \hline \begin{tabular}{c} Time-Half- \\ Life Cycles, $n$ \end{tabular} & $0 . 5 n$ & \begin{tabular}{c} Radioactive Atoms \\ (Predicted) \end{tabular} \\ \hline Initial & 1 & 100 \\ \hline 1 & 0.5 & 50 \\ \hline 2 & 0.25 & 25 \\ \hline 3 & 0.125 & 12 \\ \hline 4 & 0.0625 & 6 \\ \hline 5 & 0.03125 & 3 \\ \hline 6 & 0.015625 & 2 \\ \hline 7 & 0.0078125 & 1 \\ \hline 8 & 0.00390625 & 0 \\ \hline \end{tabular} \][/tex]
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