Radioactive decay (half life)

[OhMrsDarcy]

The problem states:

Given a sample of radioactive substance, if half of the material takes 10 years to decay into a non-radioactive material, what is the rate of decay? Given 1 ton of the material, how long until only 0.01% remains?

Our book gives the formula:

. . .A = A0(1/2)^t/h

where:

. . .A: amount at time t
. . .A0: amount at time t=0
. . .h: half life

I'm probably wrong, but the part about rate i'm guessing that it's 1/20th of the sample per year, because in 20 years it's all decayed. I've got no clue on the rest of it.

 

[stapel]

Is the formula "A = A₀(1/2)^t/h" or "A = [A₀(1/2)^t]/h"? Is "h" (the half-life variable) a unit of time? So, in this case, h = 10? How does your text define "rate of decay"?

Thank you.

Eliz.

 

[skeeter]

The problem states:

Given a sample of radioactive substance, if half of the material takes 10 years to decay into a non-radioactive material, what is the rate of decay? Given 1 ton of the material, how long until only 0.01% remains?

Our book gives the formula:

. . .A = A0(1/2)^(t/h)

where:

. . .A: amount at time t
. . .A0: amount at time t=0
. . .h: half life

I'm probably wrong, but the part about rate i'm guessing that it's 1/20th of the sample per year, because in 20 years it's all decayed. no ... in 20 years, it has decayed to 1/4 of its original amount I've got no clue on the rest of it.

A = A_o(1/2)^t/h

half life is 10 years ...

A = A_o(1/2)^t/10

01% of any original amount = .0001*A_o

.0001*A_o = A_o(1/2)^t/10

.0001 = (1/2)^t/10

use logarithms and solve for t.

 

[galactus]

Given: Half life is 10 years.

Use your equation:

\(\displaystyle \L\\\frac{A_{0}}{2}=A_{0}e^{k(10)}\)

Solve for k:

\(\displaystyle k=\frac{-ln(2)}{10}\approx{-0.069314718056}\)

Now, you have k, use it to answer the other part of the problem.

1/100th of 1% of 2000 lbs is 0.2 lbs.

 

[OhMrsDarcy]

Thanks everyone for all of your help!