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.
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
To calculate the percentage of nobelium-254 remaining at a given time, we can use the concept of half-life. The half-life of a radioactive substance is the time it takes for half of the substance to decay.
Let's start by calculating the number of half-lives for each time period:
(a) 5.0 minutes = 5.0 * 60 seconds = 300 seconds
The number of half-lives = time / half-life = 300 s / 55 s = 5.45 ≈ 5 half-lives
(b) 1.0 hour = 1.0 * 60 minutes = 60.0 minutes = 60.0 * 60 seconds = 3600 seconds
The number of half-lives = time / half-life = 3600 s / 55 s ≈ 65.45 ≈ 65 half-lives
Now, we can calculate the percentage of nobelium-254 remaining at each time:
(a) For 5.0 minutes:
Percentage remaining = (1/2)^(number of half-lives) * 100
= (1/2)^5 * 100
= (1/32) * 100
= 3.125%
(b) For 1.0 hour:
Percentage remaining = (1/2)^(number of half-lives) * 100
= (1/2)^65 * 100
= (1/36,893,488,147,419,103,232) * 100
≈ 2.71 × 10^-20 %
Therefore, the percentage of nobelium-254 remaining at 5.0 minutes after it forms is approximately 3.125%, and at 1.0 hour after it forms is approximately 2.71 × 10^-20 %.
A)
K=.693/55sec =1.26xE-2
N=e^-3.73*1E-6
N=2.28E-8
%=N/No*100=2.28%
B)
Nt=1E-6* e ^(-(.693/55)*3600 sec)
= 1.984E-26/1E-6*100
%. 2E-18%
Very small number!! I've done it 3 times?