A sample of Po-218 with a half life of 3.05 minute was allow to decay for 21.36 minutes. If there are 2.0x10^11 atoms of Po-218 remaining, how many atoms were in the original sample?
A wooden post from an ancient village has 25% of C-14 found in living trees. How old is the wooden post? T1/2 for C-14 is 5730 years.
- 翻雷滾天 風卷殘雲Lv 79 years agoFavorite Answer
1) No. of half-life elapsed = 21.36/3.05 = 7.003
So we can assume that it has elapsed for 7 half-life periods.
And no. of atoms in original sample = 2.0 x 1011 x 27 = 2.56 x 1013
2) No. of half-life periods elapsed = 2 for 25% of C-14 remaining.
Thus the age of the wooden post is 5730 x 2 = 11460 yearsSource(s): 原創答案
- 9 years ago
In fact you should put this question in the
physics category becoz radioactive decay
is actually a nuclear physics topic.
Let x be the original number
2x10^11 = x (1/2)^(21.36/3.05)
Since the amount of C-14 is 25%, it should be 1/4
as much as the original amount.
It has gone through two half lives
The age of the post = 5730x2 =11460 yesrsSource(s): Myself