+0

# little science required

+2
406
6
+420

if a radioactive element has a 5.23 days half-life and a beginning mass of 500g and has 3.90625g left. how old is the material and how many half-lives did it undergo?

Feb 22, 2018

#1
+101856
+3

OK, DC...we have this

3.90625  =  500 (1/2) ^(t/5.23)      where t is in days

Divide both sides by  500

3.90625 / 500   =  (1/2)^(t/5.23)     take the log of both sides

log  ( 3.90625 / 500)  =  log (1/2)^(t/5.23)   and we can write

log (3.90625 / 500)  =  (t / 5.23) * log (1/2)

Multiply both sides by    5.23 / log (1/2)

log (3.90625 / 500) * (5.23 / log (1/2) )  =  t  ≈ 36.61 days

To find the number of half-lives  we have

36.61 / 5.23   =  7 half-lives

Feb 22, 2018
#2
+420
+1

I am inperesed

DarkCalculis  Feb 22, 2018
#3
-1

I am inperesed that I firugered otu waht thsi wrod maens

Guest Feb 22, 2018
#4
+18449
+3

Another take (but similar)

500 e^-kt  = 250       to solve for k     we will let t= number of half lives= 1    (as in ONE half life)

k = ln (250/500) / -1    k=.693147

Now for the sample

500 e^-.693147 t = 3.90625

t = #half lifes =   ln (3.90625/5000) / - .693147     t = 7.00 half lifes  (or is it half LIVES ?)

Feb 22, 2018
#5
0

It’s “half-lives” with a hyphen. CPhil used it, but most mathematicians don’t use this because they think it means to subtract.

Part of the fun of figuring out the math problem is figuring out the written mush the answerer gives along with the solution.

Guest Feb 22, 2018
#6
+18449
0