CBSE Class 12 · 2005 · All India · Modern PhysicsShort Answer

A radioactive sample contains 2.2 mg of pure

_{6}^{11} C

, which has a half-life period of 1224 seconds. Calculate:
(i) the number of atoms present initially;
(ii) the activity when 5

μ

g of the sample is left.

Answer: $N_{0} = 1.2 \times 1 0^{20}$ ; $A \approx 1.55 \times 1 0^{14} Bq$ .

Solution

(i) Number of atoms initially:

N_{0} = \frac{m}{M} N_{A} = \frac{2.2 \times 1 0 ^{- 3}}{11} \times 6.023 \times 1 0^{23} = 1.20 \times 1 0^{20} atoms .

(ii) When

5 μ g

is left:

N = \frac{5 \times 1 0 ^{- 6}}{11} \times 6.023 \times 1 0^{23} = 2.74 \times 1 0^{17} atoms .

Decay constant

λ = \frac{0.693}{1224} = 5.66 \times 1 0^{- 4} s^{- 1}

A = λ N = 5.66 \times 1 0^{- 4} \times 2.74 \times 1 0^{17} \approx 1.55 \times 1 0^{14} disintegrations/s .

N_{0} = 1.2 \times 1 0^{20}; A \approx 1.55 \times 1 0^{14} Bq

CBSE Class 12 Board · 2005 · All India · Subjective