Radioactive Decay

[noboost4you]

After 3 days a sample of radon-222 decayed to 58% of its original amount:
a) what is the half-life of radon-222?
b) how long would it take the sample to decay to 10% of its original amount?

check my answers please:

a) m(t)=mo*e^kt
t = 3 and m(t)=.58mo
0.58mo = mo*e^3k
ln(.58) = 3k
k = -0.1816

1/2 = e^(-0.1816t)
ln(1/2)= -.1816t
t = 3.817 half life is approx 3.817 days

b) e^(-0.1816t) = .10
-0.1816t = ln(.10)
t = 12.68 it will take approx 12.68 days to reach 10%

the answer to (b) throws me. if half life is 3.817 days, 0% would be 7.634 days. where does this 12.68 days come from? i must be doing something wrong, unless (a) is wrong too.

thanks in advance

[faust9]

You have a fundamential misunderstanding of \lambda. Half-life is the time required for 1/2 of a sample to decay thus forming another product. If 1/2 of a sample decays in x days, the other half still remains... That other half would then undergo a decay for x days leaving 1/2 of the sample (1/4 of the original sample). This process continues on an on and on dividing a sample by 1/2 each time, so simply assuming 2 half lifes equals 100% of a sample is flawed.

Look here, I plugged the numbers you got into excel and spred the data out beginning from t=0\lambda to t=15\lambda. If you notice each step is approximetly 1/2 of the prevuous (this was due to in-precise rounding). It took about 11 1/2 lives after time zero for the sample to theoretically decay to an insignificant ammount which is substantially greater than 7.6 days (more along the lines of 42 days):

t0=100
t1= 50.15359446
t2= 25.15383037
t3= 12.61555008
t4= 6.327151823
t5= 3.173294066
t6= 1.591521037
t7= 0.798205007
t8= 0.400328502
t9= 0.200779133
t10=0.100697952
t11=0.050503643
t12=0.025329392
t13=0.012703601
t14=0.006371312
t15=0.003195442

[HallsofIvy]

Saying that the half-life of any substance is "T" is the same as saying that X(t)= X(0)\({\frac{1}{2}}\)^{\frac{t}{T}}

where t is the time measured in whatever units T is in.

Since it takes 3 days to decay to 58% of the original amount, we have X(3)= X(0)\({\frac{1}{2}}\)^{\frac{3}{T}}= 0.58X(0)
so that \({\frac{1}{2}}\)^{\frac{3}{T}}= 0.58
Then \frac{3}{T}ln(\frac{1}{2})= ln .58
\frac{3}{T}= 0.786
\frac{T}{3}= \frac{1}{0.786}= 1.27

and, finally T= 3(1.27)= 3.82 days. ('cuz 0.58 is pretty close to 0.5!)

Knowing now that X(3)= X(0)\({\frac{1}{2}}\)^{\frac{3}{3.82}}= 0.58X(0)
We can answer the second question by solving
\(\frac{1}{2}\)^{\frac{t}{3.82}= 0.10
\frac{t}{3.82}ln(0.5)= ln(.10)
t= 3.82(\frac{ln(.10)}{ln(0.5)})
t= 12.68 days.

"the answer to (b) throws me. if half life is 3.817 days, 0% would be 7.634 days. "

No, that would be a linear function. The whole reason this has a "half-life" is that it is an exponential function. If it decreases to half in 3.817 days, it will decrease to half of that (that is to 1/4 of the original amount) in another 3.817 days. It decreases by the same proportion not the same amount.
(Decreasing by the same amount is linear.)