The half life of phosphorus -32 is 14.3 days. What percentage of an original samples radioactivity remains after 57.2 days?
chemistry - DrBob222, Monday, May 7, 2012 at 10:29pm
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.
chemistry - Bell, Wednesday, September 23, 2015 at 11:22pm
chemistry - Bell, Wednesday, September 23, 2015 at 11:23pm