My teacher wants me to use this formula for half life question C(t) = Ce^-kt but i used a different formula plz help.
the question is:
A radioactive isotope has a half life of 30 years. If we started off with 10 mg of this isotope 12 years ago:
a) Find and expression for the amount of the isotope that is still active using time t measured in years.
b) How much of the isotope do we have at the present time?
c) What is the present instantaneous rate of decay? Show a solution using the f prime of x = the limit as h approaches 0 of f(x+h) – f(x) all over h?
d) Show a graph of the quantity of isotope over 100 years.
so, what formula did you use? was it
10*(1/2)^(t/30) ?
If so, just recall that 1/2 = e^(-ln2), so that is the same as
10e^(-ln2/30 t)
Now just plug in your various numbers.
To solve the given half-life question, let's go through the steps one by one.
a) Find an expression for the amount of the isotope that is still active using time t measured in years.
In this problem, we have a half-life of 30 years. The formula C(t) = Ce^(-kt) represents the amount of the isotope remaining at time t, where C(t) is the amount of the isotope remaining, C is the initial amount, k is the decay constant, and t is the time.
The half-life is given by the equation C(t) = (1/2)C. We can substitute these values into the formula to get:
(1/2)C = Ce^(-k*(30))
Simplifying this equation gives:
1/2 = e^(-30k)
To find the value of k, take the natural logarithm (ln) of both sides:
ln(1/2) = -30k
Solving for k:
k = ln(1/2) / -30
Therefore, the expression for the amount of the isotope remaining at time t is:
C(t) = C * e^((ln(1/2) / -30) * t)
b) How much of the isotope do we have at the present time?
To find the amount of the isotope at the present time, we substitute t = 12 years into the equation from part a:
C(12) = C * e^((ln(1/2) / -30) * 12)
c) What is the present instantaneous rate of decay?
The instantaneous rate of decay represents how quickly the isotope is decaying at the present time. It is represented by the derivative of the amount function C(t) with respect to time, denoted as C'(t).
To find the derivative, we differentiate the equation from part a:
C'(t) = (ln(1/2) / -30) * C * e^((ln(1/2) / -30) * t)
To find the present instantaneous rate of decay, substitute t = 12 years:
C'(12) = (ln(1/2) / -30) * C * e^((ln(1/2) / -30) * 12)
You can simplify this expression further if needed.
d) Show a graph of the quantity of isotope over 100 years.
To create a graph of the quantity of the isotope over 100 years, you can plot the values of C(t) for different values of t from 0 to 100 years using the equation from part a. Each point on the graph will represent the amount of the isotope remaining at a particular time.