The half life of phosphorus -32 is 14.3 days. What percentage of an original samples radioactivity remains after 57.2 days?
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To determine the percentage of an original sample's radioactivity that remains after a certain time, we need to use the concept of half-life. In this case, the half-life of phosphorus-32 is 14.3 days.
The formula to calculate the remaining percentage of the original sample's radioactivity is:
Remaining Percentage = (1/2)^(t / HL) * 100
Where:
- "t" is the time elapsed
- "HL" is the half-life
In this case, we want to find the remaining percentage after 57.2 days, so substituting the values into the formula:
Remaining Percentage = (1/2)^(57.2 / 14.3) * 100
Now, let's calculate it.
Remaining Percentage = (1/2)^(4) * 100
To evaluate the expression (1/2)^4, we start by simplifying it step by step:
(1/2)^4 = 1^4 / 2^4 = 1/16
Now we can substitute the value back into the previous equation:
Remaining Percentage = (1/16) * 100 = 6.25%
Therefore, after 57.2 days, approximately 6.25% of the original sample's radioactivity remains.
k = 0.693/t1/2/sub>
ln(No/N) = kt
No = 100 (for convenience)
N = ?
t = 57.5 days
k from above.
Then (N/No)*100 = %remaining.