posted by Michele .
A 1.00x10-6g sample of nobelium- 254, has a half-life of 55 seconds after it is formed. What is the percentage of nobelium-254 remaining at the following times?
(a) 5.0 min after it forms
(b) 1.0 h after it forms
k = 0.693/half life.
Calculate k. Then,
ln(No/N) = kt
No = 1E-6
Solve for N.
k from above.
Plug in t in seconds if youu ue 55 sec to calculte half life.
Convert to percent
Post your work if you get stuck.
Nt=1E-6* e ^(-(.693/55)*3600 sec)
Very small number!! I've done it 3 times?
I agree with your numbers. Yes, B is a small number but think about it. (a) you didn't have but a microgram initially which is small at the beginning. With a half life of 55 sec (approx 1 min) you have gone through about 60 half lives in 1 hour which leaves very little from the very little you had initially.
Thank you for your help