Solve · Q41
Q.A sample of 32P initially shows activity of one Curie. After 303 days the activity falls to 1.5 × 10^4 dps. What is the half life of 32P?
(Ans. 14.27 d)
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Start your 14-day free trial to unlock the full solution →Step 1. Convert the initial activity to dps: 1 Curie = 3.7 x 10^10 dps.
Step 2. Ratio of activities: A0/A = 3.7 x 10^10 / 1.5 x 10^4 = 2.4667 x 10^6.
Step 3. ln(2.4667 x 10^6) = ln(2.4667) + ln(10^6) = 0.9028 + 13.8155 = 14.7183.
Step 4. t1/2 = t x ln2 / ln(A0/A) = 303 x 0.6931 / 14.7183 = 210.01 / 14.7183 ≈ 14.27 d. …
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