Radioactive C14 decays according to N=No 0.5^t/HL where No is the initial amount,t in years ,HL=half life
1.if after 3506 yeasrs,65,59 percentage of C14 remains,find HL.
2.145g of C14 is left for 2029 yeasrs,how much will remain?
3.How long will it take 57g of C14 to decay to 39g?
Logarithm?
2016-12-17 10:25 pm
回答 (1)
2016-12-18 3:26 pm
✔ 最佳答案
1.65.59% * No = No * 0.5^( 3506 / HL )
0.5^( 3506 / HL ) = 65.59% = 0.6559
( 3506 / HL ) * ln 0.5 = ln 0.6559
3506 / HL = ln 0.6559 / ln 0.5
HL = 3506 * ln 0.5 / ln 0.6559 ≒ 5762
Ans: 5762 years
2.
N = 145 * 0.5^(2029/5762) ≒ 114
Ans: 114 g
3.
39 = 57 * 0.5^( t / 5762 )
0.5^( t / 5762 ) = 39 / 57
( t / 5762 ) * ln 0.5 = ln 39 - ln 57
t / 5762 = ( ln 39 - ln 57 ) / ln 0.5
t = 5762 * ( ln 39 - ln 57 ) / ln 0.5 ≒ 3155
Ans: 3155 years
收錄日期: 2021-05-02 14:11:58
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