Please help! Half-life?

Question

Find the half-life of a radioactive waste if 18 g reduces to 15 g after 5000 years?

Answer 1

Solving dy/dt=k y, y(0)=y_0 gives us

y(t)=y_0 e^(kt)

For your problem,

15=18 e^(5000 k)

so

e^k=(15/18)^(1/5000)

which means that

y(t)=18 (15/18)^(t/5000)

To find the half life, we solve the following for t:

9=18 (15/18)^(t/5000)

1/2=(15/18)^(t/5000)

ln (2)=t/5000 ln(18/15)

t=5000 ln(2)/ln (18/15) approx 19008.92

Thus, the half life is nearly 19009 years!

Note: Remember, ln (a/b)=-ln (b/a) so ln (1/2)=-ln (2); ln (15/18)=- ln (18/15).

Answer 2

You can skip the explanation and just look at the two things that have *********** beside them.

Half life problems use the continues decay model. The formula looks like this:

A = Pe^(kt). This is the exact same formula as the continuous compound growth formula.

'A' is the amount remaining. 'P' is the amount you started with. 'k' is a constant that we will have to find. 't' is the time passed in years.

We will plug the 18 and the 15 in and 5000 in and solve for 'k'. After that we will put 2 and 1 in for the starting amount and the remaining amount and solve for t.

15 = 18e^(k*5000)

15/18 = e^(k*5000)

ln(15/18) = k*5000

ln(15/18)/5000 = k ***********

Get this number off your calculator.

Let's pretend we now know what k is and do the last part. We now have the formula, but what do we plug in to it? It doesn't matter how much you start with, you know that half of it is gone. I guess it's easiest to use 2 and 1. And, it doesn't matter if you call it 2 pounds or 2 tons, because you end up dividing both sides by it. Watch:

1(pound) = 2(pound) e^(kt) (and we know k)

divide both sides by 2 pounds and you get

1/2 = e^(kt) Take the ln of both sides

ln(1/2) = kt

ln(1/2)/k = t *******

Answer 3

Use
left part of the waste = 1/2^x with x = t/T and T = 5000y
15 g / 18 g = 1/2^x = 5/6
log(1/2^x) = log(5/6)
x log(1/2) = log (5/6)
x = log (5/6) / log(1/2) = t /5000y
So t = 5000 y * log (5/6) / log(1/2)
Now grab your scientific calculator.
The answer is a number of years.

Th

Answer 4

A=A_0 2^(-t/t_0)
where A_0 is the initial amount and t_0 is tha half-life. For your situation,
15=18 2^(-5000/t_0).
Dividing by 18 and taking logarithms (base 10) on both sides gives
log(5/6)=-5000*log(2)/t_0,
so
t_0=-5000*log(2)/log(5/6).
Since log(5/6) is negative, this result turns out to be positive, as needed.
I get t_0 is about 19,000 years.

Answer 5

Well it loses one sixth after 5,000 so it would be 15,000 years
1/6 +1/6 + 1/6

Answer 6

Sounds like a job for the gravity gun.

Answer 7

19008∙92008 years.

Answer 8

is it 12000 years or is it a trick question