i have this question and im not sure how to do it i keep getting stuck with these types of questions and i would appreciate some help
(a) The rate of decay of a radioactive substance, S, is proportional to the amount remaining, x. Given that x = a at time t = 0, show that, if the time take for the amount of S to become 1/2 a is T, then at
ln(a/x)= (t/T)ln2
If T= 15 days, calculate the time taken for there to be 10% of S remaining.
ive calculated 24.14 days for the second part im not sure if this is right and i don't know how to do the first part
From the description T is the half-life of the substance
your equation becomes
x = a(2)^(-t/T)
x/a = 2(-t/T)
take ln of both sides
ln(x/a) = ln[2^(-t/T)
ln(x/a) = (-t/T)ln2
then for the second part:
.1a = a(2^(-t/15)
ln .1 = -t/15ln2
t = 49.8
check:
after 15 days .5 is left
after 30 days .25 is left
after 45 days .125 is left
my answer sounds reasonable.
can you explain how you got
x = a(2)^(-t/T)
from the information in the question
and the original equation
ln(a/x)= (t/T)ln2
not
ln(x/a) = (-t/T)ln2
which is what you had??
thanks
Didn't check for any replies til now.
in general, if you have have a doubling or halving situation, you can use a base 2, such as our equation
x = a(2)^(t/k)
if the function is increasing (doubling) then the exponent must be positive,
if the function is decreasing (like our halving) then the exponent must be negative.
Some texts and authors will change the base to (1/2) for half-life or decreasing functions, that way the exponent is always positive.
I used the first option.
Notice that using my equation the value of t comes out correctly as a positive number.
To solve the first part of the question, we need to show that ln(a/x) = (t/T)ln2.
We are given that the rate of decay of the radioactive substance, S, is proportional to the amount remaining, x. This can be represented by the differential equation:
dx/dt = kx
Where k is the proportionality constant. To solve this differential equation, we can use separation of variables:
(dx/x) = k dt
Integrating both sides gives:
ln(x) = kt + C
Where C is the constant of integration. We know that at t=0, x=a, so we can substitute these values into the equation:
ln(a) = 0 + C
C = ln(a)
Therefore, our equation becomes:
ln(x) = kt + ln(a)
To find the value of T, which is the time it takes for the amount of S to become 1/2 a, we can substitute x = a/2 and t = T into the equation:
ln(a/2) = kT + ln(a)
Next, we'll solve for k. To do this, subtract ln(a) from both sides:
ln(a/2) - ln(a) = kT
Using the logarithmic property ln(a/b) = ln(a) - ln(b), we can simplify further:
ln(a/2a) = kT
ln(1/2) = kT
Since ln(1/2) = ln(2^(-1)) = -ln(2), our equation becomes:
-ln(2) = kT
Dividing both sides by T and rearranging the terms, we have:
k = -(ln(2)/T)
Now, let's substitute this value of k back into our earlier equation:
ln(x) = [-(ln(2)/T)]t + ln(a)
Since x = a/2, we can substitute this value as well:
ln(a/2) = [-(ln(2)/T)]t + ln(a)
To simplify the equation, we can multiply both sides by -T/ln(2):
-T/ln(2) * ln(a/2) = t
Rearranging the terms gives:
ln(a/2) = (t/T)ln(2)
This matches the expression we needed to prove. Therefore, we have shown that ln(a/x) = (t/T)ln(2).
Moving on to the second part of the question, where T = 15 days and we need to calculate the time taken for there to be 10% of S remaining.
We can use the equation we derived earlier:
ln(a/2) = (t/T)ln(2)
Now, substitute a/10 for x (since 10% of S remaining means x = a/10). We can solve for t:
ln(a/10) = (t/15)ln(2)
Multiply both sides by 15/ln(2):
15 * ln(a/10) / ln(2) = t
Now, you can substitute the given value of T and calculate the time t:
t = 15 * ln(a/10) / ln(2)
To calculate t, you'll need to know the value of a.