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Sagot :
Answer:
See explanation
Explanation:
From the formula;
0.693/t1/2 = 2.303/t log (Ao/A)
t1/2 = half life of Sodium-24
Ao = initial activity of Sodium-24
A= activity of Sodium-24 at time = t
So,
0.693/15 = 2.303/15 log (800/A)
0.0462 = 0.1535 log (800/A)
0.0462/0.1535 = log (800/A)
0.3 = log (800/A)
Antilog(0.3) = (800/A)
1.995 = (800/A)
A = 800/1.995
A = 401 Bq
ii) 0.693/15 = 2.303/30 log (800/A)
0.0462 = 0.0768 log (800/A)
0.0462/0.0768 = log (800/A)
0.6 = log (800/A)
Antilog (0.6) = (800/A)
3.98 = (800/A)
A = 800/3.98
A = 201 Bq
iii)
0.693/15 = 2.303/45 log (800/A)
0.0462 = 0.0512 log (800/A)
0.0462/0.0512 = log (800/A)
0.9 = log (800/A)
Antilog (0.9) = (800/A)
7.94 = (800/A)
A = 800/7.94
A= 100.8 Bq
iv)
0.693/15 = 2.303/60 log (800/A)
0.0462 = 0.038 log (800/A)
0.0462/0.038 = log (800/A)
1.216 = log (800/A)
Antilog(1.216) = (800/A)
16.44 = (800/A)
A = 800/16.44
A = 48.66 Bq
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