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1 𝐴= 𝐴 π‘œ 𝑒 βˆ’πœ†π‘‘ LAB #2 Radioactive decay calculation ;
𝐴= 𝐴 π‘œ 𝑒 βˆ’πœ†π‘‘ πœ†= 𝑙𝑛2 𝑇 1/2 𝐴= 𝐴 π‘œ 𝑒 βˆ’ 𝑇 1/2 𝑑 Where as: A= activity at time (t) 𝐴 π‘œ = the initial activity (t=0) πœ†= decay constant t= time 𝑇 1/2 = the half-life 1 𝐢𝑖= 𝑝𝐢𝑖 1 𝐢𝑖= π‘šπΆπ‘– 1 𝐢𝑖= 10 βˆ’6 𝑀𝐢𝑖 1 𝐢𝑖= 3.7Γ— π΅π‘ž dpm 𝐢𝑖=π‘π‘’π‘Ÿπ‘– 𝑝𝐢𝑖=π‘π‘–π‘π‘œπ‘π‘’π‘Ÿπ‘–π‘’ π‘šπΆπ‘–=π‘šπ‘–π‘™π‘™π‘–π‘π‘’π‘Ÿπ‘–π‘’ 𝑀𝐢𝑖=π‘šπ‘’π‘”π‘Žπ‘π‘’π‘Ÿπ‘–π‘’ π΅π‘ž=π‘π‘’π‘π‘žπ‘’π‘’π‘Ÿπ‘’π‘™ Very Important Note:- A & 𝐴 π‘œ should have the same unit Also, t & 𝑇 1/2 should have the same unit For instant, if (A) is in mCi ( 𝐴 π‘œ ) MUST be in mCi And if (t) is in years ( 𝑇 1/2 ) MUST be in years

2 Example1 :- Given 10 mCi of 32P, which has a half-life of 14.2 days, find the quantity remaining after 60days. 𝐴= 𝐴 π‘œ 𝑒 βˆ’ 𝑇 1/2 𝑑 𝐴 0 =10π‘šπΆπ‘–, 𝐴 𝑑 =??? 𝑇 1/2 =14.2π‘‘π‘Žπ‘¦π‘  𝑑=60π‘‘π‘Žπ‘¦π‘  𝐴 𝑑 =10 𝑒 βˆ’π‘™π‘› 𝐴 𝑑 =10 𝑒 βˆ’2.93 𝐴 𝑑 =10Γ—0.0534 ∴𝐴 𝑑 =0.534 π‘šπΆπ‘– Example2 :- if the initial activity of the radionuclide mentioned above was 52 Β΅Ci, then the activity after 4 hours would be: 𝐴 0 =52¡𝐢𝑖, 𝐴 𝑑 =??? 𝑇 1/2 =14.2π‘‘π‘Žπ‘¦π‘  𝑑=4β„Žπ‘œπ‘’π‘Ÿπ‘ =0.166 π‘‘π‘Žπ‘¦π‘  𝐴 𝑑 =52 𝑒 βˆ’π‘™π‘› 𝐴 𝑑 =52 𝑒 βˆ’7.70Γ— 10 βˆ’3 𝐴 𝑑 =52Γ—0.992 𝐴 𝑑 =51.6 ¡𝐢𝑖

3 B) To how many MBq does this correspond? (99mTc Half life =6 hr)
Ex.3; A) If 20mCi of 99mTc is needed for a diagnostic test, and if 4 hours elapse between preparing of the radioisotope and its use in the test, how many mCi must be prepared? B) To how many MBq does this correspond? (99mTc Half life =6 hr) 𝐴= 𝐴 π‘œ 𝑒 βˆ’ 𝑇 1/2 𝑑 𝐴 0 =20π‘šπΆπ‘–, 𝐴 𝑑 =??? 𝑇 1/2 =6β„Žπ‘Ÿ 𝑑=βˆ’4 β„Žπ‘Ÿ A) 𝐴 𝑑 =20 𝑒 βˆ’π‘™π‘›2 6 βˆ’4 𝐴 𝑑 =20 𝑒 0.46 𝐴 𝑑 =20Γ—1.59 ∴𝐴 𝑑 =31.8 π‘šπΆπ‘– B) ∴𝐴 𝑑 =31.8 π‘šπΆπ‘–=0.0318𝐢𝑖= π΅π‘ž= π‘€π΅π‘ž

4 B) and 4 p.m. on the same day (t1/2 =6 hr) 𝐴= 𝐴 π‘œ 𝑒 βˆ’0.693 𝑇 1/2 𝑑
Ex.4; At 11 a.m. the99mTc radioactivity was measured as 9 mCi on a certain day. What was the activity at 8 a.m. B) and 4 p.m. on the same day (t1/2 =6 hr) 𝐴= 𝐴 π‘œ 𝑒 βˆ’ 𝑇 1/2 𝑑 𝐴 0 =9π‘šπΆπ‘–, 𝐴 𝑑 =??? 𝑇 1/2 =6β„Žπ‘Ÿ 𝐴) 𝑑 1 =11π‘Žπ‘š π‘‘π‘œ 8π‘Žπ‘š=βˆ’3 β„Žπ‘Ÿ B) 𝑑 2 =11π‘Žπ‘š π‘‘π‘œ 4π‘π‘š=5 β„Žπ‘Ÿ A) 𝐴 𝑑 =9 𝑒 βˆ’π‘™π‘›2 6 βˆ’3 𝐴 𝑑 =9 𝑒 0.35 𝐴 𝑑 =9Γ—1.41 ∴𝐴 𝑑 =12.7 π‘šπΆπ‘– B) 𝐴 𝑑 =9 𝑒 βˆ’π‘™π‘›2 6 5 𝐴 𝑑 =9 𝑒 βˆ’0.58 𝐴 𝑑 =9Γ—0.56 ∴𝐴 𝑑 =5.05 π‘šπΆπ‘–

5 Ex.5; What is the half-life of an isotope if it decays to 12.5% of its radioactivity in 18 minutes? 𝐴= 𝐴 π‘œ 𝑒 βˆ’ 𝑇 1/2 𝑑 𝐴 0 =100%, 𝐴 𝑑 =12.5% 𝑇 1/2 =??? 𝑑=18min 𝐴 𝐴 π‘œ = 𝑒 βˆ’ 𝑇 1/2 𝑑 𝐴 𝑑 𝐴 π‘œ = = 𝑒 βˆ’π‘™π‘›2 𝑇 1/2 18 = 𝑒 βˆ’π‘™π‘›2 𝑇 1/2 18 𝑙𝑛 =𝑙𝑛 𝑒 βˆ’π‘™π‘›2 𝑇 1/2 18 𝑙𝑛 = βˆ’π‘™π‘›2 𝑇 1/2 18 ∴ 𝑇 1/2 = βˆ’π‘™π‘›2 𝑙𝑛 12.5/ =6 π‘šπ‘–π‘›

6 Practice Problems 𝐴= 𝐴 π‘œ 𝑒 βˆ’0.693 𝑇 1/2 𝑑
𝐴= 𝐴 π‘œ 𝑒 βˆ’ 𝑇 1/2 𝑑 1. A Phosphorous-32 source has a half-life of days and had an activity of 75,000 dpm as of 12/2/92. What was the activity as of 12/28/92? 𝐴 0 =75,000 π‘‘π‘π‘š, 𝐴 𝑑 =??? 𝑇 1/2 =14.28π‘‘π‘Žπ‘¦ 𝑑=12/2/92 π‘‘π‘œ 12/28/92=26π‘‘π‘Žπ‘¦ 2. A 55mCi Sr-90 source was assayed on 6/1/88. What would the activity be on 6/1/93? (The half-life of Sr-90 is 29.1 years) H.W. 𝐴 0 =55π‘šπΆπ‘–, 𝐴 𝑑 =??? 𝑇 1/2 =29.1π‘¦π‘’π‘Žπ‘Ÿπ‘  𝑑=6/1/88 π‘‘π‘œ 6/1/93=5π‘¦π‘’π‘Žπ‘Ÿπ‘  3. An 60 mCi iodine solution of I-131 was created and then left on a laboratory shelf for two weeks before it was used. What was the activity of the solution at the time it was used? (Iodine-131 has a half-life of 8.04 days) 𝐴 0 =???, 𝐴 𝑑 =60π‘šπΆπ‘–, 𝑇 1/2 =8.04π‘‘π‘Žπ‘¦π‘  𝑑=βˆ’ 4Γ—7 =βˆ’28π‘‘π‘Žπ‘¦π‘  4. A radon air sample was collected and then counted 4 hours later. If the sample count showed an activity of 5E4 pCi, what was the activity on the sample at the time it was collected? (Radon-222 has a half-life of days) 𝐴 0 =???, 𝐴 𝑑 =5𝐸4 𝑝𝐢𝑖 𝑇 1/2 = π‘‘π‘Žπ‘¦π‘  𝑑=βˆ’ 4 24 =0.166 π‘‘π‘Žπ‘¦π‘ 

7 Practice Problems 𝐴= 𝐴 π‘œ 𝑒 βˆ’0.693 𝑇 1/2 𝑑
𝐴= 𝐴 π‘œ 𝑒 βˆ’ 𝑇 1/2 𝑑 5. A 10.5 Ci Co-60 radiography source was prepared 3 years ago, having a half-life of years. What is the activity today? 𝐴 0 =10.5 𝐢𝑖, 𝐴 𝑑 =??? 𝑇 1/2 =5.271 π‘¦π‘’π‘Žπ‘Ÿπ‘  𝑑=3 π‘¦π‘’π‘Žπ‘Ÿπ‘  βˆ’π‘‘π‘œπ‘‘π‘Žπ‘¦= π‘¦π‘’π‘Žπ‘Ÿπ‘  6. A pure alpha source reads 244,000 dpm today. It is a Po-210 source which has a half-life of days. If the source was manufactured a year ago, what the activity at the time it was manufactured? 𝐴 0 = π‘‘π‘π‘š, 𝐴 𝑑 =??? 𝑇 1/2 =138.8 π‘‘π‘Žπ‘¦π‘  𝑑=π‘‘π‘œπ‘‘π‘Žπ‘¦βˆ’1π‘¦π‘’π‘Žπ‘Ÿ π‘Žπ‘”π‘œ= π‘‘π‘Žπ‘¦π‘  7. A Cs-137 source has an activity of 750 mCi, with a half-life of years. How long will it take for the source to be read less than 100 mCi? H.W. 𝐴 0 =750π‘šπΆπ‘–, 𝐴 𝑑 =100π‘šπΆπ‘– 𝑇 1/2 =30.17 π‘¦π‘’π‘Žπ‘Ÿπ‘  𝑑=??? 8. An air sample was collected in a thorium storage building and was counted immediately, yielding 2.5E3 pCi/l. The sample was recounted 5 minutes later giving an activity of only 59.4 pCi/l. What is the half-life and the most likely isotope on the sample? H.W. 𝐴 0 =2.5𝐸3𝑝𝐢𝑖/𝑙, 𝐴 𝑑 =59.4𝑝𝐢𝑖/𝑙 𝑇 =??? 𝑑=5π‘šπ‘–π‘›π‘ 

8 A curie (Ci) is a deprecated non-SI unit of radioactivity
defined as 1 Ci = 3.7 Γ— 10¹⁰ decays per second. Therefore, 1 Ci = 3.7Β·10¹⁰ Bq (exactly) and 1 Bq = Β·10⁻¹¹ Ci. One curie is roughly the activity of 1 gram of the radium isotope ²²⁢Ra. A becquerel (Bq) is the SI derived unit of radiation activity. The Bq is defined as the activity of a quantity of radioactive material in which one nucleus decays per second. The becquerel is therefore equivalent to an inverse second, s⁻¹.


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