A mixture consists of two radioactive materials A1 and A2 with half lives of 20 s and 10 s respectively. Initially the mixture has 40 g of A1 and 160 g of A2. The amount of the two in the mixture will become equal after
- 60 s
- 80 s
- 20 s
- 40 s
Answer
(4) 40 s
Solution
Amount left after time t: N = N0 ( 12 )t/ T1/2
For equal amounts:
40 ( 12 )t/20 = 160 ( 12 )t/10
2t/10 - t/20 = 4 = 22
t20 = 2
t = 40 s
Check: after 40 s, A1 = 40 14 = 10 g, A2 = 160 116 = 10 g
Answer: (4)