更新1:
Assume Sr 90 has a half life of 28.8 years.
In an experiment, 1.6 grams of Sr-90 was buried in 1952. How much of this sample is still Sr-90 in 2008?
回答 (3)
2016-02-26 4:14 pm
Half-life of Sr-90 = 28.79 years
Time taken = 2008 - 1952 = 56 years
No. of half-lives = 56/28.79
Mass of still Sr-90 = 1.6 × (1/2)^(56/28.79) g = 0.42 g
Time taken = 2008 - 1952 = 56 years
No. of half-lives = 56/28.79
Mass of still Sr-90 = 1.6 × (1/2)^(56/28.79) g = 0.42 g
2016-02-26 4:13 pm
Roughly 0.4 g are left..
2008 - 1952 = 56. Half-life Sr-90 = c. 28 years
56 / 28 = (2 cycles)
1.6 / 2 = 0.8, 0.8 / 2 = 0.4.
2008 - 1952 = 56. Half-life Sr-90 = c. 28 years
56 / 28 = (2 cycles)
1.6 / 2 = 0.8, 0.8 / 2 = 0.4.
2016-02-26 4:10 pm
Sr-90 has a half-life of 28.78 years. So just count how many half-lives from 1952 to 2008, and divide by 2 for each half-life.
Hint: I get a number between 10% and 50%.
Hint: I get a number between 10% and 50%.
收錄日期: 2021-04-18 14:33:31
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https://hk.answers.yahoo.com/question/index?qid=20160226080425AAbtebT