Question 324058: A radioactive substance has a half-life of 6.8 hours. If 12.6 kg. remain from the original amount of 20 kg. How many hours have passed? Found 2 solutions by ankor@dixie-net.com, CharlesG2:Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! A radioactive substance has a half-life of 6.8 hours.
If 12.6 kg. remain from the original amount of 20 kg.
How many hours have passed?
:
The half-life formula: A = Ao*2^(-t/h)
Where
A = Resulting amt after t hrs
Ao = initial amt
t = time (in hrs)
h = half-life of the substance
:
20*2^(-t/6.8) = 12.6
:
2^(-t/6.8) =
2^(-t/6.8) = .63
use logs here
log(2^(-t/6.8)) = log(.63)
log equiv of exponents *log(2) = log(.63)
Find the logs of 2 and .63 *.301 = -.201 = -.201
:
-.301t = -.201 * 6.8
-.301t = -1.3645
t =
t = 4.5 hrs, for 12.6 kg to remain
:
:
Check solution on a calc; enter: 20*2^(4.5/6.8) results 12.6
You can put this solution on YOUR website! A radioactive substance has a half-life of 6.8 hours. If 12.6 kg. remain from the original amount of 20 kg. How many hours have passed?
Nt = No * (1/2)^(t/hl)
Nt = N sub t = remaining quantity after time t = 12.6 kg
No = N sub o = initial (original) quantity of a decaying substance = 20kg
t = time, in this case time in hours
hl = t sub 1/2 = half-life of the decaying quantity = 6.8 hours
(details about this equation at http://en.wikipedia.org/wiki/Half-life)
12.6 = 20 * (1/2)^(t/6.8)
12.6/20 = (1/2)^(t/6.8)
0.63 = (1/2)^(t/6.8)
logarithmic rule: if b^x = y, then logb (y) = x
log1/2 (0.63) = t/6.8
6.8 * log1/2 (0.63) = t
logarithmic rule: to change bases --> logb (x) = (logk (x))/(logk (b)),
where b is the old base and k is the new base
log1/2 (0.63) = (log10 (0.63))/(log10 (1/2)) = 0.666576266274808225391191653851317 = approx. 0.666576 to 6 places
t = 6.8 * log1/2 (0.63) = 4.53271861066869593266010324618895
t = approximately 4.5 hours has passed
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