Finding the half-life of an unknown substance

[Sam]

1000 particles of an unknown radioisotope decays to 472.37 particles in 50 days.

(a) What is the half-life of this substance?


Any problems that I had done previously, the half-life was given.

Any help will be greatly appreciated.

Well, I know a few things, but don't know if they apply to this problem:

The decay constant = .693/T base 1/2
t 1/2 = half-life

Time to decay to a single particle = LN(Nbase0)/decay constant

N = number of particles remaining at some elapsed time
Nbase0 = number of particles we started with
e = a symbol that represents the irrational number 2.718281828...
lambda = decay constant
x = elapsed time

The formula to determing how much of the substance will be remaining at any particular time is: N = Nbase0 e^-lambda x

[arcnets]

Hi Sam,
you give the correct formulae
N = N0e-[lamb]x
and
[lamb]=.693/T1/2.
You could combine these, solve for T1/2, and plug in N0, N, and x.

Or, more instructive, you could use the equation
N = N0 * (1/2)x/T1/2.

In both cases, the important step is applying the ln() to both sides of the equation.

[Sam]

Thank you.

I'll give it a try.

[Sam]

I used, based on the above:

N(t) = N base0^(-lambda x /T),
where T is the half-life
t = 50 days
N base0 = 1000

I rounded to the nearest whole number for days and came up with 46.

Is that correct?

[arcnets]

Originally posted by Sam
I used, based on the above:

N(t) = N base0^(-lambda x /T)
I think it should read
N(x) = N0eln(1/2)x/T, but that's probably just a typo, since
T is the half-life
t = 50 days
N base0 = 1000
is correct (except x=50 days), and your answer is also correct, although they probably expect you to come up with some more decimals...[;)]

[Sam]

I have two more parts to this problem:

(a) How much will be left after 75 days?
(b) How long will it take the 1000 particles to decay to a single particle?

For (a), I came up with: 324.656 particles
For (b), I came up with: 460.521 days

Will you please verify my answers?

Thank you!

[arcnets]

Both of your answers are correct.