Question

Radium 226 has a decay rate that satisfies the differential equation D'(t)= -.00043D(t). Assume that D(0)=100.

What is D(t)?

How much is left after 100 years?(round to nearest hundredth.)

To the nearest year, what is the life half life of Radium 226?

Answer 1

The decay rate of Radium 226 can be described by the differential equation D'(t) = -0.00043D(t). Solving this equation with the given initial condition, we find the function D(t) = 100e^(-0.00043t). After 100 years, approximately 28.02 units of Radium 226 remain. The half-life of Radium 226 is approximately 1609 years.

Step-by-step explanation:

The decay rate of Radium 226 can be modeled by the differential equation D'(t) = -0.00043D(t), where D(t) represents the amount of Radium 226 at time t. To find the function D(t), we can solve this differential equation. We can rewrite it as D'(t)/D(t) = -0.00043 and integrate both sides. This gives us ln(D(t)) = -0.00043t + C, where C is the constant of integration. Using the initial condition D(0) = 100, we can solve for C and find that C = ln(100). Therefore, the function D(t) = 100e^(-0.00043t).

Now, to find the amount of Radium 226 after 100 years, we can substitute t = 100 into the function D(t). D(100) = 100e^(-0.00043*100), which is approximately equal to 28.02. Rounded to the nearest hundredth, the amount of Radium 226 left after 100 years is 28.02.

To find the half-life of Radium 226, we need to determine the time it takes for half of the initial amount to decay. Half of the initial amount is 100/2 = 50. Using the function D(t) = 100e^(-0.00043t), we can solve for t when D(t) = 50. 50 = 100e^(-0.00043t), which simplifies to e^(-0.00043t) = 0.5. Taking the natural logarithm of both sides, we have -0.00043t = ln(0.5), and solving for t gives t = ln(0.5)/(-0.00043). Using a calculator, this is approximately equal to 1609.54. Rounded to the nearest year, the half-life of Radium 226 is 1609 years.

Answer 2

A)
`D(t)=100e^(-0.00043t)`

B)
`{D(t=100)}=95.8`

C) τ=2326 years

Step-by-step explanation:

A)

Consider the differential equation:

`(dD)/(dt)=-0.00043D`

Separate variables dD and dt:

`(dD)/(D)=-0.00043dt`

Integrate: Remember that The integral of dD/D is ln(D).

`\int\limits^(D)_(D_0) {(1)/(D) } \, dD=\int\limits^(t)_(0) {-0.00043} \, dt`

`ln((D)/(D_0))=-0.00043t\\(D)/(D_0)=e^(-0.00043t)`

`D=D_0*e^(-0.00043t)`

And do
`D_0=100`, so:

`D=100*e^(-0.00043t)`

B)

The function we have found is a function that let us know how many of Ra is after t years, so, evaluate the function at t=100:

`D(t=100)=100*e^(-0.00043*100)=95.791\\`

Which rounded is 95.8.

C) For this equation it is known that the life half life τ is equal to (1/λ), where λ is the exponential number that is multiplying to -t in the D(t) function; so for this case, λ=0.00043, and the life half life is
`(1)/(0.00043)=2325.58`,

which rounded is equal to 2326 years.