## Trending News

# Radioactive decay

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.

### 2 Answers

- 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 years

Source(s): 原創答案- Login to reply the answers

- 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)

x =2.57x10^13

Since the amount of C-14 is 25%, it should be 1/4

as much as the original amount.

1/4=(1/2)^2

It has gone through two half lives

The age of the post = 5730x2 =11460 yesrs

Source(s): Myself- Login to reply the answers