Q: The half life of At215 is 100 μ sec. If a simple contains 215 mg of At215, the activity of the sample initially is
(a) 102 Bq
(b) 3 × 10 Bq
(c) 4.17 × 1024 Bq
(d) 1·16 × 105 Bq
Click to See Answer :
Ans: (c)
Sol : T = 100 µs = 10-4 s
λ = 0.6931/T
= 0.6931/(10-4)
= 0.6931 × 104 s-1
No. of atoms is 215 mg of sample
$ \displaystyle \frac{6.023 \times 10^{23}}{215}\times 215 \times 10^{-3}$
N = 6.023 × 1020
Activity, $\displaystyle \frac{dN}{dt} = \lambda N $
$ \displaystyle \frac{dN}{dt} = 0.6931 \times 10^4 \times 6.023 \times 10^{20} $
= 4.17 × 1024 Bq