Grace asked in 科學及數學化學 · 9 years ago

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

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    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
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  • 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
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