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

Answer:

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:

$\frac{dD}{dt}=-0.00043D$

Separate variables dD and dt:

$\frac{dD}{D}=-0.00043dt$

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

$\int\limits^{D}_{D_0} {\frac{1}{D} } \, dD=\int\limits^{t}_{0} {-0.00043} \, dt$

$ln(\frac{D}{D_0})=-0.00043t\\\frac{D}{D_0}=e^{-0.00043t}$

$D=D_0*e^{-0.00043t}$

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 $\frac{1}{0.00043}=2325.58$,

which rounded is equal to 2326 years.