Asked by matt

The half life of phosphorus -32 is 14.3 days. What percentage of an original samples radioactivity remains after 57.2 days?
Answered by DrBob222
k = 0.693/t<sub>1/2/sub>
ln(No/N) = kt
No = 100 (for convenience)
N = ?
t = 57.5 days
k from above.
Then (N/No)*100 = %remaining.
Answered by Bell
Poooty butt
Answered by Bell
potty butt
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