Ncert Exercises & Solutions

13.10 The half-life of \({ }_{38}^{90} \mathrm{Sr}\) is 28 years. What is the disintegration rate of 15 mg of this isotope?

Hint: The rate of disintegration is given by: \(\frac{\mathrm{dN}}{\mathrm{dt}}=-\lambda \mathrm{N}\).

Step 1: Find the number of atoms present in 15 mg of \({ }_{38}^{90} \mathrm{Sr}\).

The half-life of

{}_{38}^{90}{Sr}, t_{1 / 2} = 28

years

= 28 \times 365 \times 24 \times 60 \times 60

= 8.83 \times 10^{8} s

Mass of the isotope, m = 15 mg

90 g of {}_{38}^{90}

Sr atom contains

6.023 \times 10^{23}

(Avogadro's number) atoms.

Therefore, 15 mg of

{}_{38}^{90}{Sr}

contains atoms

= \frac{6.023 \times 10^{23} \times 15 \times 10^{- 3}}{90} = 1.0038 \times 10^{20}

Step 2: Find the rate of integration of \({ }_{38}^{90} \mathrm{Sr}\).
Rate of disintegration,

\frac{dN}{dt} = λN

Where,

λ =

decay constant

= \frac{0.693}{8.83 \times 10^{8}} S^{- 1}

Therefore,

\frac{dN}{dt} = \frac{0.693 \times 1.0038 \times 10^{20}}{8.83 \times 10^{8}} = 7.878 \times 10^{10}

atoms / s

Hence, the disintegration rate of 15 mg of the given isotope is 7.878x

10^{10}

atoms/ s.