Chemistry Question?

Question

27. 64^Co decays by a first-order process via the emission of a beta particle. The 64^Co isotope has a half-life of 7.8 min. How long will it take for 31/321 of the colbalt to undergo decay??

a. 7.8 min
b. 16 min.
c. 23 min
d. 31 min
e. 39 min

Answer 1

You have typed the question wrong, and you meant to type 31/32.
This takes 5 x half lives, which is 39 minutes.

Answer 2

Since the half life of Co^64 is 7.8 minutes, that means that in 7.8 minutes, only half of the Co^64 will remain. But you're looking for 31/321, which is kind of weird number, but roughly equal to 10%, right? So in two half lives, there will be 1/4 remaining (1/2 x 1/2), and that will take 15.6 minutes (just 7.8 x 2). That gets us to 25%. So let's go another half life (3 so far). That gets us to 12.5% remaining, and will take 23.4 minutes. Another half life gets us to 6.25% and would take 31.2 minutes. But now we've gone too far, since we only want rougly 10% to remain, or 31/321. So let's look at our choices. We know that a) is wrong, since that's only one half life and gets us to 50%. Since we know that 3 half lives was pretty close, getting us to 12.5%, we know it should be close to this, or 23.4 minutes. We also know that 4 half lives was overboard - that got us to 6.25%, and took over 31 minutes. So it looks like c.) 23 minutes is pretty darn close to what we're looking for, and by the process of elimination, must be the right answer.
Good luck!
Just looked at Gervald's post, and he's probaby right. So if you mis-typed it, listen to his answer! You can use the same logic though. Good luck!