posted by Madison .
Will you check this answer for me?
the half-life of strontium-87 is 2.8hr. What percentage of this isotope will remain after 8 hours and 24 minutes?
I'd be interested in knowing how you came up with that answer.
8 hr and 24 min, according to my calculation is [(8*60)+24/60] = 8.4 hrs.
k = 0.693/t1/2 = 0.693/2.8 = 0.2475
ln(No/N) = kt
I took No = 100 (an arbitrary number)
Solved for N
k from above
t = 8.4
Solving for N, I obtained 12.5 so if we started with 100 g at time = zero, then at the end of 8.4 hours we would have 12.5 g left. %left = (12.5/100)*100 = ??
Check my work.