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Sagot :
To predict the pattern for the decay of 100 atoms over the course of eight half-life cycles and fill in the table, follow these steps:
1. Understand the Half-Life Process: Every half-life cycle, the number of atoms is reduced by half.
2. Calculate the Number of Atoms Remaining After Each Half-Life: Use the formula:
[tex]\[ \text{Remaining Atoms} = 100 \times (0.5)^n \][/tex]
where [tex]\( n \)[/tex] is the number of half-life cycles.
Let's fill in the values step-by-step and record the remaining atoms rounded to the nearest whole number:
- Initial (0 half-life cycles):
[tex]\[ \text{Remaining Atoms} = 100 \times (0.5)^0 = 100 \][/tex]
- After 1 half-life cycle (n = 1):
[tex]\[ \text{Remaining Atoms} = 100 \times (0.5)^1 = 50 \][/tex]
So, [tex]\( A = 50 \)[/tex].
- After 2 half-life cycles (n = 2):
[tex]\[ \text{Remaining Atoms} = 100 \times (0.5)^2 = 25 \][/tex]
(Already given in the table)
- After 3 half-life cycles (n = 3):
[tex]\[ \text{Remaining Atoms} = 100 \times (0.5)^3 = 12.5 \][/tex]
Round 12.5 to the nearest whole number:
[tex]\[ B = 12 \][/tex]
- After 4 half-life cycles (n = 4):
[tex]\[ \text{Remaining Atoms} = 100 \times (0.5)^4 = 6.25 \][/tex]
(Already given in the table)
- After 5 half-life cycles (n = 5):
[tex]\[ \text{Remaining Atoms} = 100 \times (0.5)^5 = 3.125 \][/tex]
Round 3.125 to the nearest whole number:
[tex]\[ C = 3 \][/tex]
- After 6 half-life cycles (n = 6):
[tex]\[ \text{Remaining Atoms} = 100 \times (0.5)^6 = 1.5625 \][/tex]
Round 1.5625 to the nearest whole number:
[tex]\[ D = 2 \][/tex]
- After 7 half-life cycles (n = 7):
[tex]\[ \text{Remaining Atoms} = 100 \times (0.5)^7 = 0.78125 \][/tex]
Round 0.78125 to the nearest whole number:
(Already given in the table as 1)
- After 8 half-life cycles (n = 8):
[tex]\[ \text{Remaining Atoms} = 100 \times (0.5)^8 = 0.390625 \][/tex]
Round 0.390625 to the nearest whole number:
[tex]\[ E = 0 \][/tex]
Now, we complete the table:
[tex]\[ \begin{tabular}{|c|c|c|} \hline \begin{tabular}{c} Time-Half- \\ Life Cycles, $n$ \end{tabular} & $0.5^{\boldsymbol{n}}$ & \begin{tabular}{c} Radioactive Atoms \\ (Predicted) \end{tabular} \\ \hline Initial & 1 & 100 \\ \hline 1 & 0.5 & 50 \\ \hline 2 & 0.25 & 25 \\ \hline 3 & 0.125 & 12 \\ \hline 4 & 0.0625 & 6 \\ \hline 5 & 0.03125 & 3 \\ \hline 6 & 0.015625 & 2 \\ \hline 7 & 0.0078125 & 1 \\ \hline 8 & 0.00390625 & 0 \\ \hline \end{tabular} \][/tex]
Therefore, the values are:
[tex]\[ A = 50 \][/tex]
[tex]\[ B = 12 \][/tex]
[tex]\[ C = 3 \][/tex]
[tex]\[ D = 2 \][/tex]
[tex]\[ E = 0 \][/tex]
1. Understand the Half-Life Process: Every half-life cycle, the number of atoms is reduced by half.
2. Calculate the Number of Atoms Remaining After Each Half-Life: Use the formula:
[tex]\[ \text{Remaining Atoms} = 100 \times (0.5)^n \][/tex]
where [tex]\( n \)[/tex] is the number of half-life cycles.
Let's fill in the values step-by-step and record the remaining atoms rounded to the nearest whole number:
- Initial (0 half-life cycles):
[tex]\[ \text{Remaining Atoms} = 100 \times (0.5)^0 = 100 \][/tex]
- After 1 half-life cycle (n = 1):
[tex]\[ \text{Remaining Atoms} = 100 \times (0.5)^1 = 50 \][/tex]
So, [tex]\( A = 50 \)[/tex].
- After 2 half-life cycles (n = 2):
[tex]\[ \text{Remaining Atoms} = 100 \times (0.5)^2 = 25 \][/tex]
(Already given in the table)
- After 3 half-life cycles (n = 3):
[tex]\[ \text{Remaining Atoms} = 100 \times (0.5)^3 = 12.5 \][/tex]
Round 12.5 to the nearest whole number:
[tex]\[ B = 12 \][/tex]
- After 4 half-life cycles (n = 4):
[tex]\[ \text{Remaining Atoms} = 100 \times (0.5)^4 = 6.25 \][/tex]
(Already given in the table)
- After 5 half-life cycles (n = 5):
[tex]\[ \text{Remaining Atoms} = 100 \times (0.5)^5 = 3.125 \][/tex]
Round 3.125 to the nearest whole number:
[tex]\[ C = 3 \][/tex]
- After 6 half-life cycles (n = 6):
[tex]\[ \text{Remaining Atoms} = 100 \times (0.5)^6 = 1.5625 \][/tex]
Round 1.5625 to the nearest whole number:
[tex]\[ D = 2 \][/tex]
- After 7 half-life cycles (n = 7):
[tex]\[ \text{Remaining Atoms} = 100 \times (0.5)^7 = 0.78125 \][/tex]
Round 0.78125 to the nearest whole number:
(Already given in the table as 1)
- After 8 half-life cycles (n = 8):
[tex]\[ \text{Remaining Atoms} = 100 \times (0.5)^8 = 0.390625 \][/tex]
Round 0.390625 to the nearest whole number:
[tex]\[ E = 0 \][/tex]
Now, we complete the table:
[tex]\[ \begin{tabular}{|c|c|c|} \hline \begin{tabular}{c} Time-Half- \\ Life Cycles, $n$ \end{tabular} & $0.5^{\boldsymbol{n}}$ & \begin{tabular}{c} Radioactive Atoms \\ (Predicted) \end{tabular} \\ \hline Initial & 1 & 100 \\ \hline 1 & 0.5 & 50 \\ \hline 2 & 0.25 & 25 \\ \hline 3 & 0.125 & 12 \\ \hline 4 & 0.0625 & 6 \\ \hline 5 & 0.03125 & 3 \\ \hline 6 & 0.015625 & 2 \\ \hline 7 & 0.0078125 & 1 \\ \hline 8 & 0.00390625 & 0 \\ \hline \end{tabular} \][/tex]
Therefore, the values are:
[tex]\[ A = 50 \][/tex]
[tex]\[ B = 12 \][/tex]
[tex]\[ C = 3 \][/tex]
[tex]\[ D = 2 \][/tex]
[tex]\[ E = 0 \][/tex]
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