Question

The half life of ${}_{38}^{90}Sr$ is 28 years. The disintegration rate of 15 mg of this isotope is of the order of.

• A. $10^{11}Bq$
• B. $10^{10}Bq$
• C. $10^{7}Bq$
• D. $10^{9}Bq$

Answer 1

Answer: (B)

$10^{10}Bq$

Answer 2

: Here, $T_{1/2} = 28years = 28 \times 3.154 \times 10^{7}s$

As number of atoms in 90 g of ${}_{38}^{90}sr$ $= 6.023 \times 10^{23}$

$\therefore$number of atoms in 15 mg of ${}_{38}^{90}sr$

$= \frac{6.023 \times 10^{23}}{90} \times \frac{15}{1000}$

i.e. $N = 1.0038 \times 10^{20}$

rate of disintegration, $= \frac{0.693}{T_{1/2}}N$

$= \frac{0.693 \times 1.0038 \times 10^{20}}{28 \times 3.15 \times 10^{7}} = 7.877 \times 10^{10}Bq = 10^{10}Bq$