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Sagot :
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]
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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