Please help with this radioactive question!

Cobalt 60 which undergoes beta decay has a half life of 5.26 years How many beta particles are emitted in 600s by a 3.75 mg sample of cobalt 60?

find K for the reaction
1/2 life = 0.693 / k
0.693 / 5.26yrs = k = 0.132

[A] / [Aº] = e^(-kt)
[A] = 3.75mg x e(-0.132 x 1.9x10^-5yrs)
[A] = 3.7499mg Co left

9.4x10^6mg lost to decay
9.4x10^-9 g / 60g/mole = 1.57x10^-7moles atoms

1.57x10^-7 x 6.022x10^23 = 9.44x10^16 atoms decayed
9.44x10^16 emissions

In one half-life, half of the 3.75 mg will decay. That's 1.875 mg. Each of those atoms emitted a beta particle. I'm going to carry several guard digits below and then round off at the end.

Let's see how many beta particles that was:

0.001875 g / 59.9338 g/mol = 0.0312845 mol of Co-60

0.0312845 mol times 6.022 x 10^23 atoms/mol = 1.88395 x 10^22 atoms

So, now we know how many beta particles were emitted in 5.26 years.

I use this:

https://secure.wikimedia.org/wikipedia/e…

to get the atomic weight of Co-60

Now, I want to see what fraction 600 seconds is of one year. I'm going to convert 5.26 yr to seconds using Google:

1.65989 x 10^8 seconds / one half-life

Now, I want to see what fraction 600 seconds is of one half-life

600 / 1.65989 x 10^8 = 3.614697 x 10^-6

Final calc:

1.88395 x 10^22 beta emissions/half life times 3.614697 x 10^-6 half-life = 6.8099 x 10^16 emissions

Three sig figs seems reasonable:

6.81 x 10^16 emissions

Lots of calculations, so double-check everything I did. Email me if you find I made a mistake. Thanks.

Nice problem.

Update:

found this Co-60 problem:

http://answers.yahoo.com/question/index?…

It looks pretty interesting too.