A radioactive element decays exponentially according to the function Q(t) = Q0ekt
.If 100 mg of the element decays to 25 mg in 12 days, find the half-life of the element.
Did not need the attitude bob pursley
Plug in the values given from the problem into Q(t) = Q0e^kt (Q(12) = 25, Q0 = 100, t = 12) and then solve for k
25 = 100e^(12k)
ln(1/4) = lne^(12k)
ln(1/4) = 12k
ln(1/4)/12 = k
Next, set Q(t) = 50 (since 50 is half of 100) and plug in the previously found value of k to solve for t
50 = 100e^(ln(1/4)/12)t
1/2 = e^(ln(1/4)/12)t
ln(1/2) = (ln(1/4)/12)t
12ln(1/2) = ln(1/4)t
12ln(1/2)/ln(1/4) = t
t = 6
The half-life of the element is 6
Well, radioactive decay is serious business, but why not add a little humor to the equation? Let's solve this radioactive riddle together!
We're given that Q(t) = Q0e^kt, where Q(t) is the amount of radioactive element at time t, Q0 is the initial amount of the element, k is the decay constant, and e is Euler's number (approximately 2.71828).
We're also given that after 12 days, the amount of the element decreased from 100 mg to 25 mg. So, we can set up the equation:
25 = 100e^(12k)
To find the half-life, we need to find the time it takes for the amount of the element to decrease by half.
Let's say the initial amount is Q0, and after the half-life, it becomes Q0/2. We can set up the equation:
Q0/2 = Q0e^(k*t_half)
Dividing these two equations, we get:
(25 / 100) = (Q0e^(12k)) / (Q0e^(k*t_half))
Simplifying, we have:
1/4 = e^(11k*t_half)
Now we can take the natural logarithm of both sides to solve for k*t_half:
ln(1/4) = 11k*t_half
ln(1/4) = -ln(4)
1/4 = 4
11k*t_half = ln(4)
t_half = ln(4) / (11k)
Well, t_half is the half-life we're trying to find! We just need to calculate the value of k.
To do that, we can use the fact that after 12 days, the amount of the element decreased from 100 mg to 25 mg:
25 = 100e^(12k)
Dividing by 100, we get:
1/4 = e^(12k)
Now, let's take the natural logarithm of both sides:
ln(1/4) = ln(e^(12k))
ln(1/4) = 12k
k = ln(1/4) / 12
Now we can substitute the value of k into our equation for the half-life:
t_half = ln(4) / (11 * (ln(1/4) / 12))
Simplifying further, we have:
t_half = 12ln(4) / (11ln(1/4))
Using a calculator to evaluate this expression, we find:
t_half ≈ 18.48 days
So, our radioactive riddle has been solved! The half-life of the element is approximately 18.48 days.
To find the half-life of the radioactive element, we need to determine the amount of time it takes for half of the initial amount to decay.
Given:
Initial amount (Q0) = 100 mg
Amount after decay (Q(t)) = 25 mg
Time (t) = 12 days
The radioactive decay function is given by:
Q(t) = Q0 * e^(kt)
To find the value of k, we can use the given information. Substituting the values into the exponential decay equation:
25 = 100 * e^(k * 12)
To isolate the variable k, we can divide both sides of the equation by 100:
0.25 = e^(12k)
To solve for k, take the natural logarithm (ln) of both sides:
ln(0.25) = ln(e^(12k))
Using the property of logarithms, ln(e^x) = x, we can simplify further:
ln(0.25) = 12k
Now, divide both sides of the equation by 12:
k = ln(0.25) / 12
Using a calculator, we find that k is approximately -0.0575.
The half-life (T) is the amount of time it takes for the initial quantity to decrease by half. We need to find the value of T when the decay constant (k) is known.
The half-life (T) can be calculated using the formula:
T = ln(2) /|k|
Substituting the value of k into the equation:
T = ln(2) / |-0.0575|
Using a calculator, we find that T is approximately 12.04 days.
Therefore, the approximate half-life of the radioactive element is 12.04 days.
story problem? Calculus is not elementary school.
Q(t)=Qo*ekt
25=100*e12k
ln of both sides...
ln(1/4)=12k
-1.38629436=k*12
solve for k
then, for half life...
.5=ek*t
solve for t (take ln of both sides)
thalf=ln(.5)/k