Hey, can somebody here help me with this stuff?

Question

1) Why are radioisotopes with long half-lives not administered internally in medical procedures?

2) What % of C-14 would you expect a piece of 34000 year-old fossilized bone from a mastodon to have when compared to a similar piece of bone from a modern elephant?

1. I thought the answer would be obvious. If you were administered a dose of a radioactive isotope would you rather have one that was gone in 3 days or one that stayed with you for the next 50 years (being bombarded with disintegration products for all of that 50 years)?

2.
k = 0.693/t_1/2
Find the value of C-14 half-life, substitute into the equation above and calculate k.
Then, use ln(No/N)=kt.
You know k and t (t=34000 yrs). Start with 100 atoms for No and solve for N, the number of atoms today. Compare that with 100 to obtain the ratio.
Post your work if you get stuck and need further assistance.

1) That's what I thought. It just seemed too simple of an answer.

2) So I'd do this?:

k=.693/5730 years
k=.0001
In(100/N)=.0001*t
Would t be 5730*2?

What is ln(No/N)=kt?

We never learned that.


2) So I'd do this?:

k=.693/5730 years
k=.0001 You dropped a 21 here. It should be 0.000121
In(100/N)=.0001*t
Would t be 5730*2? No. t is 34000 in the problem. So it would be
ln(100/N)=0.000121*34000 and solve for N. Then you can do the percent. By the way, that is not an In, it is Ln but the L is written in lower case to be ln.

Your answer to #1 is correct. That didn't seem to post.

I also forgot to answer your other question. The equation
ln(No/N)=kt where k=0.693/t_1/2 is the first order equation for radioactive decay. No=atoms at the start, N=atoms at the finish, k is the constant which depends upon the half-life and is calculated separately as I have shown, and t is the time spent in going from atoms at the start to atoms at the finish.