A radioactive sample contains 2.2 mg of pure ${}_{6}^{11}C,$ which has half-life period of 1224 seconds.
Calculate
(i) the number of atoms present initially.
(ii) the activity when 5 μg of the sample will be left.

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Solution

Given,
Mass of pure sample of carbon,

{}_{6}^{11}C

= 2.2 mg
Half-life period of sample, T_1/2 = 1224 seconds

(i) The number of atoms present intially is,

= \frac{2.2 \times 10^{- 3} \times 6.023 \times 10^{23}}{11} = 1.2 \times 10^{20} atoms

(ii) The number of atoms present in 5 μg,

= \frac{5 \times 10^{- 6} \times 6.023 \times 10^{23}}{11} = 2.74 \times 10^{17} atoms

Activity is given by,

R = λN = \frac{0.693 N}{T_{1 / 2}} = \frac{0.693 \times 2.74 \times 10^{17}}{1224} = 1.55 \times 10^{14} .

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