更新1:
Half lives: Ar= 12 min Pb=37 min K=22 min Indium=43 min
How long would it take (min) for argon-44 to decrease from 96.0 to 12.0mg? Ar-44 Pb-196 K-44 Indium-117?
回答 (2)
2017-11-27 7:33 am
From 96.0 to 12.0 mg :
12.0/96.0 = 1/8 = (1/2)³
Hence, time taken = 3 half-lives
Time taken for Ar-44 = (12 min) × 3 = 36 min
Time taken for Pb-196 = (37 min) × 3= 111 min
Time taken for K-44 = (22 min) × 3 = 66 min
Time taken for indium-117 = (43 min) × 3 = 129 min
12.0/96.0 = 1/8 = (1/2)³
Hence, time taken = 3 half-lives
Time taken for Ar-44 = (12 min) × 3 = 36 min
Time taken for Pb-196 = (37 min) × 3= 111 min
Time taken for K-44 = (22 min) × 3 = 66 min
Time taken for indium-117 = (43 min) × 3 = 129 min
2017-11-27 7:13 am
(12.0 mg / 96.0 mg) = (1/2)^(z / 12 min)
log (12/96) = (z / 12 min) log (1/2)
log (12/96) / (log (1/2)) = z / 12 min
z = (12 min) log (12/96) / (log (1/2)) = 36 minutes
log (12/96) = (z / 12 min) log (1/2)
log (12/96) / (log (1/2)) = z / 12 min
z = (12 min) log (12/96) / (log (1/2)) = 36 minutes
收錄日期: 2021-04-18 18:04:32
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20171126230424AAJ34Km