Answer:
500.3 Bq
Explanation:
From the formula;
0.693/t1/2 = 2.303/t log (Ao/A)
Where;
t1/12 = half life = 5 years
Ao = initial activity = 2000 Bq
0.693/5 = 2.303/10 log (2000/A)
0.1386= 0.2303 log (2000/A)
0.1386/0.2303 = log (2000/A)
0.6018 = log (2000/A)
(2000/A) = Antilog (0.6018)
(2000/A) = 3.9976
A = 2000/3.9976
A = 500.3 Bq