radium decomposes at a rate proportional to the amount instantaneously present.Find the half-life of radium if three-fourths of it is present after 8 hours
If y = f(x) then you have
dy/dx = ky
dy/y = k dx
ln y = kx + ln c
y = c e^(kx)
c is the initial amount present (e^0 = 1)
Now, e^(ln 1/2) = 1/2, so we can fold that into the k value, giving
y = c (1/2)^(kx)
since we want to find the half-life.
y(8)/y(0) = (1/2)^(8k) = 3/4
8k = ln(3/4)/ln(1/2)
k = 0.05188 = 1/19.27
so y = c (1/2)^(x/19.27)
and the half-life is 19.27 hours
where does the y(8)/y(0) also 8k comes from?
huh? That is the fraction left after 8 hours!
y(0) is the original amount
y(8) is the amount at time x=8
i got it awhile ago, thank u for answering!
y(8)/y(0) = (1/2)^(8k) = 3/4
especially this part y(8)/y(0) does it need to be in fraction? is it just y only since it is y= ce^kx
i kinda don't get this part but i know that there's a given 3/4 can u please elaborate it i just wanna know and learn tnx.
To find the half-life of radium given that three-fourths of it is present after 8 hours, we can use the concept of exponential decay.
Let's denote the initial amount of radium as "A0" and the amount of radium present after time "t" as "A(t)". According to the given information, three-fourths of radium is present after 8 hours, which means that A(t) = (3/4) * A0.
The rate at which radium decomposes is proportional to the amount present, which can be expressed as:
dA/dt = -k * A(t)
Where "k" is the proportionality constant.
We can solve this differential equation to find the general form of A(t). Integrating both sides with respect to t, we get:
∫ (1 / A(t)) dA = -k * ∫ dt
This simplifies to:
ln(A(t)) = -k * t + C
Where "C" is the constant of integration.
Since A(t) = (3/4) * A0 and t = 8 hours when three-fourths of radium is present, we can substitute these values into the equation to solve for "C":
ln((3/4) * A0) = -k * 8 + C
Now, let's find the half-life, which is the time it takes for half of the initial amount of radium to decay. Half of the initial amount corresponds to A(t) = (1/2) * A0.
We can rewrite the general equation with A(t) = (1/2) * A0 and solve for the half-life:
ln((1/2) * A0) = -k * t_half + C
Since ln((1/2) * A0) = ln(A0) - ln(2) and ln(A0) = -k * 8 + C, we can substitute these values into the equation:
ln(1/2) + ln(A0) = -k * t_half + (-k * 8 + C)
ln(1/2) = -k * t_half
Using ln(1/2) = -0.693 (approximately), we can solve for t_half:
-0.693 = -k * t_half
Simplifying, we have:
t_half = 0.693 / k
Therefore, the half-life of radium is equal to 0.693 divided by the proportionality constant "k".