Q: The half life of At^{215} is 100 μ sec. If a simple contains 215 mg of At^{215}, the activity of the sample initially is

(a) 10^{2} Bq

(b) 3 × ^{10} Bq

(c) 4.17 × 10^{24} Bq

(d) 1·16 × 10^{5} Bq

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Ans: (c)

Sol : T = 100 µs = 10

^{-4}sλ = 0.6931/T

= 0.6931/(10^{-4})

= 0.6931 × 10^{4} 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 × 10^{20}

Activity, $\displaystyle \frac{dN}{dt} = \lambda N $

$ \displaystyle \frac{dN}{dt} = 0.6931 \times 10^4 \times 6.023 \times 10^{20} $

= 4.17 × 10^{24} Bq